I have watched more than 20 videos on this topic and asked my professor and none of them know how to explain how to find the limits of integration or even attempt to explain it. Thank the heavens for this video and this teacher. Thank you so much.
To everyone asking why he took half of sin squared: it is the FORMULA for finding the area of a polar curve. You take the integral of 1/2 r^2, r being your function of theta. If the equation is r = cos theta you would do the integral of 1/2 cos^2 theta. If you graph the trig function as a function of x it helps to see where its maximums, zeros, and minimums occur and where it's increasing and decreasing. Since sin has a period of 2pi, sin of 4theta has a period of pi/2, with a maximum value occurring at pi/8 (sin would normally have a maximum value of 1 at pi/2 but since it's sin4theta everything happens 4 times faster). If you don't get that, just make a value chart, pick values of theta to get your r and graph the thing, stop after you finish one loop and figure out what angle interval it is
adding something to itsKyyyle's tip- the formula 1/2y^2dtheta is derived from area of triangle , just like you use rectangles in cartesian coordinate system . suppose f(x) is at some angle theta initially and moves a small angular distance dtheta . this will form a sector looking shape but since dtheta is really small , the arc can be taken as a straight line. area = 1/2.base.height = 1/2.y.y.dtheta = 1/2.y^2dtheta. there u go , integrate it
WithASideOfFries 'twas the eve of his funeral, for he had a short life to live. Battered and bruised by his calculus exam, he had nothing else to give. Rip Tim schott
I saw Josh's comment and it's a great idea. After I graduate and get a job, I am going to send you money. Thanks for helping us all. You saved my life.
Totally just learned everything I need to know about Conics and Polar calc from your videos. Which is nice, because I have a test in 20 minutes. Pure. God. Lifesaver.
dude, no joke you're awesome. this video is 10 times more powerful then hastily taken notes in class. if i get confused a can just come back and watch it again as reference :D
What's up Patrick, Its taking me much longer to get through a single one of your videos because I find myself pausing it, getting up, and doing a little dance every time I understand something better. Shoot, I almost did a head spin when you cleared up how to find the limits of integration. I was killing myself over this. U the man!
I have a great math teacher this semester, seriously top notch, unlike most commenters on here. but these vids have helped me a lot in seeing the material and steps again, sometimes with slightly different techniques from what I learned in class. I like your straightforward style of teaching. very useful stuff. I've sent links to these vids to other students in my class as well. unlike most people I love math, but I still need help sometimes and these vids have been the help I've needed.
Thanks so much Patrick. I feel somewhat ok about taking my final tomorrow thanks to your vids. I've understood more in these past few days than I did all semester! If I pass my calc 2 class, it's all thanks to you!! Really do appreciate all your hard work!!
misanthropic or not, your contribution towards students around the world is not only globally renown, but globally respected. I studied abroad in the middle east for 2 semesters, and I can tell you that you were watched there. As thousands have said before me: Thank you.
you explain the basic concepts of what you are covering first, clearly and easy, then u go through guided examples just as easily WHY CANT OUR TEACHERS UNDRESTAND THAT
@patrickJMT Please don't take it in the wrong way I didn't mean anything bad by it. I really love your videos. You quite literally have been my professor for my calc 2 class. My prof is horrible. I know people complain about their mediocre profs but you have no idea how bad mine is. He talks like the guy from the clear eyes commercial, he writes directly in front of him so we cant see what hes doing, and he talks to the board as he teaches. I really appreciate all your videos. Thanks A LOT!
Wow, thank you soooo much, this video made graphing polar curves so easy to me. I haven't understood it for weeks (my teacher made it way more confusing), so thanks SOO much!
We want to find the area in ONE loop of the graph (in the instructions). If you plot points you will see exactly why he chose 0 and pi/4. He was just making an argument that each time sin (4theta) = 0 we will have gone through one loop of the graph. If you go further to 2(pi)/4 you will see that we make a second loop etc...Hope that helps.
Okay I don't want to sound stupid and you probably answered this in the video but I couldn't find it so why can't you just take the integral from 0 to tau (2pi) instead of solving for when sin(4*theta) = 0?
You could do that, but then you would end up with the combined area of 8 loops because the graph of the polar function sin(4*theta) is swept out through the entire unit circle if you integrated from 0 to Tau. Since the question only asks for one loop, the limits of integration can be just the first two angles where r=0: 0 and pi/4.
Thanks for having these videos, I really wish I knew about you before this semester. I plan on retaking calc 2 and I will be returning next semester for your videos. On my way to failing the exam ='(
thank you for publishing your videos..i am in math 208 or calc2 and it is helping me a lot.. i would like you to do more complicated problems... i could not find a harder one could you tell me where do i have to look for it? and u are right the most complicated thing is to find the limits i stck w them each time thank you
i have my Maths 1111 test for tommorow and i did not know a damn thing about integration and polar coordiantes until i watched your videos! thanks a lot...at least maybe i won't be failing!
I'm a student @ Dal I have an A plus in a course I almost dropped until I found your videos. Honestly Patrick; Is there anything I can do to thank you?
thanks Patrick, this was helpful. Does anyone know where the next video is, the "more complicated" one? Looking to find area between two curves in polar.
Great example! Thanks for all your help. But in the future can you make your examples somewhat more difficult. I'm in calc 2 and problems on our exams are never this easy. However you help tremendously... THANK YOU!!!
well, here i am going through your videos again for my math final tomorrow! deja vu :o anyway i feel like this is just an overly repeated statement from other comments i read but its just THAT true: how thankful i am for your videos and that if it wasn't for you i may have just dropped out of calculus :S keep on making math seem cool 8)
Patrick, I know your good at math and science but what else can you do? How about a video of you dancing or something? Anyway, thanks for making these videos.
Patrick, in this case where u only find the area of one loop, the limits are o ad pi/4. But say we were to find the area of the whole thing? I how would i know what to times that mess of an integral with? pls reply soon, my final is tom!!!!! im stuck on this topic!!
ReverseJam1 The question is asking to find the area of one loop of the function. From the graph, you can see that from 0 to pi/4 is a whole loop. It goes from radius of 0 at theta = 0 to another radius of 0 at theta = pi/4, and since pi/8 will give the max radius, it means that the graph would have to go back from 0 to p/4 in a circular-ish motion with p/8 as the maximum r value in the loop, which would give one enclosed loop.
+Paulo Herce Im not sure if your allowed to use a calculator but if you change your mode from FUNC to POL and then enter the equation you can see the graph itself. This makes it a lot easier for me.
Plug in the angle into the function and you'll notice that at the angle pi/8, sin(4theta) = sin(4pi/8) = sin(pi/2) = 1, and since sin(x) is less than or equal to 1 for all x, at the angle pi/8 the function sin(4theta) reaches 1, which is the maximum value.
its really suprising that you consider yourself having misanthrope tendencies seeing how you help thousands of students on a daily basis with your videos
Thanks. Having though trouble with finding these types "r = 1 + 2 sin(theta) & r = 3 + 2 cos (theta). I don't know how to find the limits of integrations with these equations. Any help appreciated
Hi and thanks for the video. I watched the video because I was looking for a proof of the formula which I didn't find it here. Can you please explain it? Or give a reference to it? Thanks.
@patrickJMT He also doesn't finish the problems, he writes all over the place on the board, he forgets to tell us when the homework is due, and he gives us the wrong page numbers for the HW and still expects us to turn it in on the next class session. And he also makes the test a lot harder than what we do in the book and in class. Its bad man.
No, it's because integral of cos4θ is sin4θ/4 and when you plug π/4 you get sinπ/4 which is 0. However if you would use cos2θ after integration you end up with sinπ/2 which is 1. Hope it helps, your case was just coincidence
To check your answers for petal problems: given r=c*sin(n theta) Area of one petal = c^2*(pi/4n) so r=5sin(7 theta) Area of one petal = 5^2*[pi/(4*7)] = 25*pi / 28
LOL I used to teach high school math, but calculus never came up. I'm a bookkeeper these days, so nope, I have no use for all of this nonsense. It was fun at the time, though. Math is a young man's game.
Oh. I misinterpreted the question. I thought that by "one loop" it means the entire way round, like one revolution = one loop. I was thinking that for some unknown reason the equation didn't work when there are zero values, and I also thought he forgot to multiply by 8 at the end. Thanks for clearing that up.
for the function r^2=2sin(3theta), should you also divide by 2 to get the limits? The solution shows the limits for the function to be (0, pi/6). This problem was a part of my problem set last semester for calculus II and this semester for calculus III and its killing me that I dont get it.
Thanks for the video, it's helped me quite a bit. However, I have run into a strange problem which I cannot solve; Find the area bounded by the polar graph of r^2 = 4cos(2theta) I know that I just do 1/2 times the integral of r^2 but I cannot find the correct limits. The answer keeps coming out to be 0, which is wrong because I've looked up the graph. If you have time, could you please help me out? I'd be very grateful! (Not saying that I'm not grateful already! :D)
I have watched more than 20 videos on this topic and asked my professor and none of them know how to explain how to find the limits of integration or even attempt to explain it. Thank the heavens for this video and this teacher. Thank you so much.
You're one of the few people that knows this stuff very well and can explain it in a way everyone understands.
Good job, man.
To everyone asking why he took half of sin squared: it is the FORMULA for finding the area of a polar curve. You take the integral of 1/2 r^2, r being your function of theta. If the equation is r = cos theta you would do the integral of 1/2 cos^2 theta. If you graph the trig function as a function of x it helps to see where its maximums, zeros, and minimums occur and where it's increasing and decreasing. Since sin has a period of 2pi, sin of 4theta has a period of pi/2, with a maximum value occurring at pi/8 (sin would normally have a maximum value of 1 at pi/2 but since it's sin4theta everything happens 4 times faster). If you don't get that, just make a value chart, pick values of theta to get your r and graph the thing, stop after you finish one loop and figure out what angle interval it is
adding something to itsKyyyle's tip- the formula 1/2y^2dtheta is derived from area of triangle , just like you use rectangles in cartesian coordinate system . suppose f(x) is at some angle theta initially and moves a small angular distance dtheta . this will form a sector looking shape but since dtheta is really small , the arc can be taken as a straight line. area = 1/2.base.height = 1/2.y.y.dtheta = 1/2.y^2dtheta. there u go , integrate it
This is super late but Very helpful
pi over 4 is just chilling.
i giggled at that
twas the eve of the eve of the calculus bc exam...
+tim schott How did it go for you? How did you do?
WithASideOfFries 'twas the eve of his funeral, for he had a short life to live. Battered and bruised by his calculus exam, he had nothing else to give. Rip Tim schott
He fought well...
rip, i feels his pain now
He gone
I saw Josh's comment and it's a great idea. After I graduate and get a job, I am going to send you money. Thanks for helping us all. You saved my life.
this guy should receive the BRObel Prize
+Josh Vanselow 10/10 flawless victory
How I met your mother
more like how i met your step-mother*
Totally just learned everything I need to know about Conics and Polar calc from your videos. Which is nice, because I have a test in 20 minutes. Pure. God. Lifesaver.
8 years later and you are still saving lives.
Someday when I'm a successful engineer I'm going to send you money.
any luck yet?
lol he failed the exam
Josh Vanselow elon musk yet
4 years later. Are you an Engineer now?
its been 4 yrs, say something.
@EnchantedEnchantress glad you like the videos : ) thanks for the kind words!
dude, no joke you're awesome. this video is 10 times more powerful then hastily taken notes in class. if i get confused a can just come back and watch it again as reference :D
What's up Patrick, Its taking me much longer to get through a single one of your videos because I find myself pausing it, getting up, and doing a little dance every time I understand something better. Shoot, I almost did a head spin when you cleared up how to find the limits of integration. I was killing myself over this. U the man!
I have a great math teacher this semester, seriously top notch, unlike most commenters on here. but these vids have helped me a lot in seeing the material and steps again, sometimes with slightly different techniques from what I learned in class. I like your straightforward style of teaching. very useful stuff. I've sent links to these vids to other students in my class as well. unlike most people I love math, but I still need help sometimes and these vids have been the help I've needed.
Thanks so much Patrick. I feel somewhat ok about taking my final tomorrow thanks to your vids. I've understood more in these past few days than I did all semester! If I pass my calc 2 class, it's all thanks to you!! Really do appreciate all your hard work!!
misanthropic or not, your contribution towards students around the world is not only globally renown, but globally respected. I studied abroad in the middle east for 2 semesters, and I can tell you that you were watched there. As thousands have said before me: Thank you.
amazing. I *NEVER* understood how to find the angle theta of a problem like this until today. Thank you, thank you, thank you!!
you explain the basic concepts of what you are covering first, clearly and easy, then u go through guided examples just as easily
WHY CANT OUR TEACHERS UNDRESTAND THAT
Feels like old youtube was worth to watch...all the videos of initial years are 100x better than of today's ....
You are the man! I use your videos so many times in Calc 2
@patrickJMT Please don't take it in the wrong way I didn't mean anything bad by it. I really love your videos. You quite literally have been my professor for my calc 2 class. My prof is horrible. I know people complain about their mediocre profs but you have no idea how bad mine is. He talks like the guy from the clear eyes commercial, he writes directly in front of him so we cant see what hes doing, and he talks to the board as he teaches. I really appreciate all your videos. Thanks A LOT!
thanks, i have a quiz over area of polar graphs in 4 hours and this really helped.
Thanks for posting these videos. They really help. Plain and simple.
I hope you get big bucks from these videos because you definitely deserve it.
i wouldn't call them 'big' :)
That's what she said Pat...
i love you. Thank you for this. Finding the limits of integration was what I was stuck on.
@fizzo68 spread the word about the videos is the best 'thank you' i can get!
Wonderful walk through to find the area of a petal. Thank you!
Wow, thank you soooo much, this video made graphing polar curves so easy to me. I haven't understood it for weeks (my teacher made it way more confusing), so thanks SOO much!
actually , i am in your class
you made it much easier to understand
Thanks from Sweden
@denokindo it is my pleasure
We want to find the area in ONE loop of the graph (in the instructions). If you plot points you will see exactly why he chose 0 and pi/4. He was just making an argument that each time sin (4theta) = 0 we will have gone through one loop of the graph. If you go further to 2(pi)/4 you will see that we make a second loop etc...Hope that helps.
@DrunkinHulk spring break was last week!
@Gilster85 that is what i see, at least
Great videos.Really appreciate you taking time to make them and share your knowledge.
Anyone else who found it amusing when he said, "pi over four is just chilling..." lol
Thank you so much for helping us out.
@omarofuae well, i am very flattered!
@patrickJMT I still love you Patrick and the all the hardwork you have done for us Calculus newbs =)
Okay I don't want to sound stupid and you probably answered this in the video but I couldn't find it so why can't you just take the integral from 0 to tau (2pi) instead of solving for when sin(4*theta) = 0?
YOU MADE THIS SO EASY. THANK YOU!!!!!!
excellent!
Now, you are one of my heroes Patrick :))
You could do that, but then you would end up with the combined area of 8 loops because the graph of the polar function sin(4*theta) is swept out through the entire unit circle if you integrated from 0 to Tau. Since the question only asks for one loop, the limits of integration can be just the first two angles where r=0:
0 and pi/4.
Thank you so much for doing this! You saved me last semester with Taylor and Power Series! Now this! Much appreciated(:
i wish you put more videos on this subject up. like finding the inner loop area of a limacon. and other common polar conic shapes.
Thanks for having these videos, I really wish I knew about you before this semester. I plan on retaking calc 2 and I will be returning next semester for your videos. On my way to failing the exam ='(
Im just learning this stuff and your video really makes sense. Thanks for that
What kind of camera do you use? Your videos are in great resolution and amazing quality. Please provide the model and type. : )
thank you for publishing your videos..i am in math 208 or calc2 and it is helping me a lot.. i would like you to do more complicated problems... i could not find a harder one could you tell me where do i have to look for it? and u are right the most complicated thing is to find the limits i stck w them each time thank you
i have my Maths 1111 test for tommorow and i did not know a damn thing about integration and polar coordiantes until i watched your videos!
thanks a lot...at least maybe i won't be failing!
This is the smartest video I've ever seen which contains "just chillin'"
I'm a student @ Dal I have an A plus in a course I almost dropped until I found your videos. Honestly Patrick; Is there anything I can do to thank you?
thanks Patrick, this was helpful. Does anyone know where the next video is, the "more complicated" one? Looking to find area between two curves in polar.
Great example! Thanks for all your help. But in the future can you make your examples somewhat more difficult. I'm in calc 2 and problems on our exams are never this easy. However you help tremendously... THANK YOU!!!
well, here i am going through your videos again for my math final tomorrow!
deja vu :o
anyway i feel like this is just an overly repeated statement from other comments i read but its just THAT true: how thankful i am for your videos and that if it wasn't for you i may have just dropped out of calculus :S
keep on making math seem cool 8)
thank you sooooo much. your videos help me survive calc :)
Patrick, I know your good at math and science but what else can you do? How about a video of you dancing or something? Anyway, thanks for making these videos.
Another great video Patrick. Your the man!
I really hope you have multivariable calc videos, as well. Calc 3 next semester, YAY!!!
You're a life saver! These videos help a lot! :)
I can't seem to find the "more complicated" follow-up video that you mention at the end of this video. Do you have a link?
Patrick,
in this case where u only find the area of one loop, the limits are o ad pi/4. But say we were to find the area of the whole thing? I how would i know what to times that mess of an integral with?
pls reply soon, my final is tom!!!!!
im stuck on this topic!!
sir you just saved my life.
i did not get it how you get the limits?
me too, he skipped this part very quickly for sure
ReverseJam1
The question is asking to find the area of one loop of the function. From the graph, you can see that from 0 to pi/4 is a whole loop. It goes from radius of 0 at theta = 0 to another radius of 0 at theta = pi/4, and since pi/8 will give the max radius, it means that the graph would have to go back from 0 to p/4 in a circular-ish motion with p/8 as the maximum r value in the loop, which would give one enclosed loop.
+Paulo Herce Im not sure if your allowed to use a calculator but if you change your mode from FUNC to POL and then enter the equation you can see the graph itself. This makes it a lot easier for me.
How does one know that pi/8 will give you the max radius? How would one determine the point at which you obtain the max radius?
Plug in the angle into the function and you'll notice that at the angle pi/8, sin(4theta) = sin(4pi/8) = sin(pi/2) = 1, and since sin(x) is less than or equal to 1 for all x, at the angle pi/8 the function sin(4theta) reaches 1, which is the maximum value.
Thank you sooo much! This makes much more sense now.
knowing there is 8 loops, could you just solve from 0 to 2pi and multiply by 1/8?
did you ever find out if you could just multiply by 1/8?
its really suprising that you consider yourself having misanthrope tendencies seeing how you help thousands of students on a daily basis with your videos
Thanks. Having though trouble with finding these types "r = 1 + 2 sin(theta) & r = 3 + 2 cos (theta). I don't know how to find the limits of integrations with these equations. Any help appreciated
Hi and thanks for the video. I watched the video because I was looking for a proof of the formula which I didn't find it here. Can you please explain it? Or give a reference to it?
Thanks.
i frequent one of the thunderbird coffee shops .... :)
hey. great video. would it be possible for you to make a follow-up video with more complicated problems?
Whats the name of your other video that shows you how to find the area between curves in polar? you mentioned it in this video and I can't find it.
@patrickJMT He also doesn't finish the problems, he writes all over the place on the board, he forgets to tell us when the homework is due, and he gives us the wrong page numbers for the HW and still expects us to turn it in on the next class session. And he also makes the test a lot harder than what we do in the book and in class. Its bad man.
Could you do one for area between common interiors?
yea, it does not exist yet... i will try to do one later today
it is a great town, i love it here. and unfortunately, my misanthrope tendencies do not allow for strangers in the house
Any idea how to calculate volume of revolution for sin4x?
Where can I find the second video he was speaking of? The area between two polar curves.
Can do an example of finding the limits of integration when their is no coefficient for theta
you are so awesome! Would you do more problem on drawing the polar curve ex: 2cos0+1 . I don't know how to graph the inner loop. thanks so much!
No, it's because integral of cos4θ is sin4θ/4 and when you plug π/4 you get sinπ/4 which is 0.
However if you would use cos2θ after integration you end up with sinπ/2 which is 1. Hope it helps, your case was just coincidence
Can you please send me the link of the other complicated example, I cannot find it.
To check your answers for petal problems:
given r=c*sin(n theta)
Area of one petal = c^2*(pi/4n)
so r=5sin(7 theta)
Area of one petal = 5^2*[pi/(4*7)] = 25*pi / 28
is there a formala for cos as well?
@@asinner I don't remember. I don't remember any of this stuff.
The fact that you replied lmao. It’s alright. At this stage in your life, I doubt you need any of this.
LOL I used to teach high school math, but calculus never came up. I'm a bookkeeper these days, so nope, I have no use for all of this nonsense.
It was fun at the time, though. Math is a young man's game.
Hi. this is Mandar. I have question about this.
Why you are considering only half of the square of function????
how did you go from 1/2 (1-cos8theta)dtheta to 1/4 [0-sin8theta/8)?
thx man, this helps out a lot.
Oh. I misinterpreted the question. I thought that by "one loop" it means the entire way round, like one revolution = one loop. I was thinking that for some unknown reason the equation didn't work when there are zero values, and I also thought he forgot to multiply by 8 at the end. Thanks for clearing that up.
why did you square the sin(4theta) and divide it by 2 when the question only had sin(4theta) at 1:25
for the function r^2=2sin(3theta), should you also divide by 2 to get the limits? The solution shows the limits for the function to be (0, pi/6). This problem was a part of my problem set last semester for calculus II and this semester for calculus III and its killing me that I dont get it.
thank you sooo much you are the best i loveeee you, you helped me alot thank you thank you thank you
6:48 Inside the bracket: Sin8Theta over 8 where this denominator 8 came from? Thanks.
Since you used u-substitution, don't you have to change the bounds of the integral?
Amanda Kenney you don’t change the limits when you use a trig identity
sir
how to find the area of region outside the circle r=2 and inside the lemniscate r^2 = 8 cos2theta and what is meant by lemniscate
Can I ask is there any relationship between the formula of centroid's y coordinate and this one? Looks the same to me....
thank you so mach in spite of i from iraq but i indersand every think,,,,,,,,,
Duha Duha it's amazing you're learning on your own! Keep going :)
Thanks for the video, it's helped me quite a bit. However, I have run into a strange problem which I cannot solve;
Find the area bounded by the polar graph of r^2 = 4cos(2theta)
I know that I just do 1/2 times the integral of r^2 but I cannot find the correct limits. The answer keeps coming out to be 0, which is wrong because I've looked up the graph. If you have time, could you please help me out? I'd be very grateful! (Not saying that I'm not grateful already! :D)
Thanks a lot man you are AWESOME!
I understand this problem... I don't understand how to find the interval of integration when the graph looks very similar to a circle... Plze help!!
But where does that formula come from in the first place? How do you derive the area formula?
it is in any standard calculus textbook
yes he did search (area between curve)
Can I donate to you? Do you have a donation site?