It's insane how closely your video matches the material my class is going over. I literally got a test problem that was the same as your example problem in the video. thanks
I used to use khan academy but wow, this is more in-depth with more examples. Thank you so much. Honestly this is where I should be paying my tuition. Thanks Again.
Professor Organic Chemistry Tutor, thank you for an incredible video/lecture on Finding the Area Bounded by Two Polar Curves in Calculus Two. Graphing the two Polar Curves is an important first step when finding the Area between two Polar Curves in Calculus. When I took Calculus many years ago, I have problems with this material. This is an error free video/lecture on UA-cam TV with the Organic Chemistry Tutor.
Thank you so much, 13:11 where you decide on the a and b boundaries I was confused on why, so I did a little experimenting and realized that a=5pi/3 would never get to a b=pi/3 in the positive counterclockwise direction because it is increasing. You would have to adjust it to b=2pi+pi/3=7pi/3 and then you do get the right answer of pi or you can do what you did, but I would be careful with this hack in the case where the shape of the area is not symmetric and doubling would not work.
Yes, in the case of it not being symmetric, you could integrate from -π/3 to π/3 or any angle you may choose. This works because you're only increasing in your angle.
Your videos are my favorite to use for help with subjects math and physics always find answers here and you break them down nicely thank you and keep up the great work!
For the first example (r=3cosx, r=1+cosx), I am getting a different answer. I am getting A= pi - sqt(3)/2... because my calculator is resolving -2sin(pi/3)= -2sqrt(3)/2, and sin(2(pi/3))= sqrt(3)/2... which means they can't cancel out. I also checked on Symbolab and Mathway calculators to make sure my calculator wasn't miscalculating the results. This would change the final answer, yes?
You need to trace the graph. See at the rightmost point of the graph the angle is zero then trace big circle till 2π/3, then trace the small circle, now you need to set theta=180 degrees in r=1+2cos(theta) which gives us -1. And remember r is the distance from origin to the curve. So, that means when we rotate by π, the value of the distance of the curve from origin should be 1 and that should be opposite from the direction we are rotating due to the minus sign.(You can also check it like this => In 2π/3 to 4π/3 the value of r is always negative that means the curve won't be there in 2π/3 to 4π/3 region). So that's why we are starting from π as the small circle cuts the x axis after we have rotated by π. Now trace the upper part of the small circle. Now set r=0 as we can see the small circle is passing through origin so the distance from origin to the graph is zero(We get theta=2π/3 and 4π/3). That means we were at π and we need to go till 4π/3 to get the complete area of the small circle above x axis. That's why the limits are from π to 4π/3. Someone might think why aren't we taking the limits as 0 to 2π/3 for the small circle as well. But if we do that you will get => 0->2π/3 f(x)dx - 0->2π/3 f(x)dx = 0. Which doesn't make any sense cause we can clearly see the curve is definitely covering some area.
17:53 Great video overall, although this is not the correct answer why have you disregarded the fact that the inside of 2cos(2theta) is equivalent to 4cos(theta)... instead it was canceled with the subtracted 2cos(theta) term which makes absolutely no sense.
I understand this is a calculus 2 video, but who can explain how solving the problem this way rather than the algebraic or geometry (or visual, finding shapes) methods? As I'm typing, I'm realizing that it's the one size fits all idea if you get what I mean... Anyways
when you integrate cos(2theta), you need to put a 1/2 in front of the sin. think of substituting u = 2theta so that you have cos(u). since u = 2theta, du = 2, which you will have to equalize by putting a 1/2 out front. if you integrate, it will become 1/2sin(u). then plug u back in and it becomes 1/2sin(2theta). however, be careful when you're using u substitution with limits because then you will have to adjust your upper and lower bounds.
every point from pi to 4pi/3 is actually placed in the first quadrant. This happens because if theta=pi in r=1+2costheta, r will be negative, and since we are talking about polar coordinates it suffers a 180 degrees turn, going to the first quadrant.
Can someone please tell me what I should do if I'll get a negative area? Is it okay or not? I did almost everything already but I can't seem to get my area to be positive... My brain's already dying
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It's insane how closely your video matches the material my class is going over. I literally got a test problem that was the same as your example problem in the video. thanks
they both may be using same book
Lucky you.
Your teacher has probably watched his video
@@3110-l6n shiiii
Same here
I used to use khan academy but wow, this is more in-depth with more examples. Thank you so much. Honestly this is where I should be paying my tuition. Thanks Again.
Professor Organic Chemistry Tutor, thank you for an incredible video/lecture on Finding the Area Bounded by Two Polar Curves in Calculus Two. Graphing the two Polar Curves is an important first step when finding the Area between two Polar Curves in Calculus. When I took Calculus many years ago, I have problems with this material. This is an error free video/lecture on UA-cam TV with the Organic Chemistry Tutor.
You’re the best teacher in the world! God bless you! ❤️
Thank you so much, 13:11 where you decide on the a and b boundaries I was confused on why, so I did a little experimenting and realized that a=5pi/3 would never get to a b=pi/3 in the positive counterclockwise direction because it is increasing. You would have to adjust it to b=2pi+pi/3=7pi/3 and then you do get the right answer of pi or you can do what you did, but I would be careful with this hack in the case where the shape of the area is not symmetric and doubling would not work.
Yes, in the case of it not being symmetric, you could integrate from -π/3 to π/3 or any angle you may choose. This works because you're only increasing in your angle.
Your videos are my favorite to use for help with subjects math and physics always find answers here and you break them down nicely thank you and keep up the great work!
he just jumps right in. i love it
This is actually really relaxing to listen to while also teaching math
Thanks. Was worried about this but feeling good about it now
Thanks a lot. I have been learning from you since trig.
I have been watching him since trig as well!
@@Bee-uy2cnwhat are yall up to these days
You are the savoir of modern engineering students.
This video is so clear, it covers every possible situation for calculating polar area! Keep up!
Extremely helpful. The step by step is great.
It seems its a bit of a nightmare to actually get the exact answer...?
For sure it is😭😭😭😭
usually you’re allowed calculator when you do polar areas (if you’re in calc bc), so you can just plug and go without having to go to math hell
@@SeaguIISoup bro I have my calc final tomorrow and on every sample exam there’s these types of questions and I get no calculator
@@jordanyupyup3120 my condolences
@@SeaguIISoup I just hope I passed bro I really do
Thank you alot..l want you make videoes about vectors & matrix pleaae
Keep up the good work bro, your videos really help me man
THANK YOU I HAVE TRANSCENDED
i swear bro you saved my life lil nigga i could not understand this stuff for so long but you came in clutch. May God bless you
This made this so clear to me thank you :)
For the first example (r=3cosx, r=1+cosx), I am getting a different answer. I am getting A= pi - sqt(3)/2... because my calculator is resolving -2sin(pi/3)= -2sqrt(3)/2, and sin(2(pi/3))= sqrt(3)/2... which means they can't cancel out. I also checked on Symbolab and Mathway calculators to make sure my calculator wasn't miscalculating the results. This would change the final answer, yes?
I am seeing what I did now... it's 2sin(2(pi/3)). D'oh!
this comment may bring back memories
@@oktayozdemir2617 it definitely did. Haha I am graduated now. Lol
Why can't you take the limit from 0 to 2pi and then minus the integral with limits 2pi/3 to 4pi/3?
yo why did this show up on my recommendations how did youtube know I needed this???
Can you use symmetry on all of the examples in the video??
A great man indeed 👏 thanks bro
When integrate from π/3 to 5π/3.the answer is 2π.is my answer right?
i also get that
Could you have had the limits of integration from. (-pi/3,pi/3)?
Such amazing video!! For real, you saved me in this subject, it is the first explanation that was visual, direct and made me want to learn more.
Hello there
Ø 300mm R = 250mm 90 °
Can give 5 pieces of elbow area, excel formula.
thanks
I wish he would drop some cool new merch! I would cop that drip immediately
22:15 Why have you taken limits from PI... Shouldnt it be taken from 0??
Thank you so much ❤
22:20 why you not take the limit in 2nd integral from 2π/3 to 4π/3?
i think its because he is only calculating one half of region not whole small circle
yeah cuz i think taking the limit would probably have to be reserved for the improper integrals, and here we’re making definite integrals so yeah
what goes on inside the integral if the question is to find area of the region that lies inside of both curves.
Honestly i can't see why we need to use pi in the integration to 4pi/3 at 22:13. Anyone can enlighten me?
You need to trace the graph. See at the rightmost point of the graph the angle is zero then trace big circle
till 2π/3, then trace the small circle, now you need to set theta=180 degrees in r=1+2cos(theta) which gives us -1. And remember r is the distance from origin to the curve. So, that means when we rotate by π, the value of the distance of the curve from origin should be 1 and that should be opposite from the direction we are rotating due to the minus sign.(You can also check it like this => In 2π/3 to 4π/3 the value of r is always negative that means the curve won't be there in 2π/3 to 4π/3 region). So that's why we are starting from π as the small circle cuts the x axis after we have rotated by π. Now trace the upper part of the small circle. Now set r=0 as we can see the small circle is passing through origin so the distance from origin to the graph is zero(We get theta=2π/3 and 4π/3). That means we were at π and we need to go till 4π/3 to get the complete area of the small circle above x axis. That's why the limits are from π to 4π/3.
Someone might think why aren't we taking the limits as 0 to 2π/3 for the small circle as well. But if we do that you will get =>
0->2π/3 f(x)dx - 0->2π/3 f(x)dx = 0. Which doesn't make any sense cause we can clearly see the curve is definitely covering some area.
Appreciate your work
hello! in the first part of the video how did you get the 5pi/6? it is 1/2 but how did you know its 5pi/6??
18:39
what happens if we integrate from 0 to 2pi ? we will get area for what ?
The whole curve.
@@strangerdanger7616 too late bro 😂😂 thanks anyway 🙌🏻
It took him 10 minutes to solve the first problem damn Calculus 2
bro saved my calc grades
excellent vid! thanks for your efforts!
thank you ☺
why are you not changing the sign when you are deriving cos theta?
17:53 Great video overall, although this is not the correct answer why have you disregarded the fact that the inside of 2cos(2theta) is equivalent to 4cos(theta)... instead it was canceled with the subtracted 2cos(theta) term which makes absolutely no sense.
Thanks man u are really helpful
Thanks
I understand this is a calculus 2 video, but who can explain how solving the problem this way rather than the algebraic or geometry (or visual, finding shapes) methods? As I'm typing, I'm realizing that it's the one size fits all idea if you get what I mean... Anyways
God bless you
18:12 i think the final answer is 2pi because you forgot to double your answer LOL good work though helps alot
yup
u noticed also :(
no the ans is only pi, because he cancel out 2 & 1/2 from the formula at first...
When integrate from π/3 to 5π/3.the answer is 2π.is my answer right?
@@shafinsarwar6448 then why when i do it from 5pi/3 to pi/3, I get 2pi
how do you know how to distinguish between r1 and r2
As you go radially outward, the first curve you see will be your r2
Could u plz tell me how to get the shaded region of the curves r=1 and r=2sintheta
Isn’t the first theta in problem three 5pi/4. Cosx and sinx have different values at theta = 0
What topic that you get costheta=1/2 = pie/3 and 5pie/4 in 12:04
I'm not sure enough but I think the answer for the last question is wrong, and correct me if I'm wrong but I think it should be pi/8..
how do you graph those equations
Dis man da 🐐
6:45 why did you divide by 2 after u integrate? i'm the darkest brain in my class.... but sure i'll get there.
when you integrate cos(2theta), you need to put a 1/2 in front of the sin. think of substituting u = 2theta so that you have cos(u). since u = 2theta, du = 2, which you will have to equalize by putting a 1/2 out front. if you integrate, it will become 1/2sin(u). then plug u back in and it becomes 1/2sin(2theta). however, be careful when you're using u substitution with limits because then you will have to adjust your upper and lower bounds.
A god amongst men
nobodys a god leave it out mate..
Thankkssss
What grade does this lesson have?
I know I'm replying 6 years later but this is taught in University Calculus 2.
I don't get how the upper half of the inner curve for 22:04 starts at pi...assiste moi si vu plait??
every point from pi to 4pi/3 is actually placed in the first quadrant. This happens because if theta=pi in r=1+2costheta, r will be negative, and since we are talking about polar coordinates it suffers a 180 degrees turn, going to the first quadrant.
I also didn't understand
At 22:10, how does one determine that the inner loop starts at pi?
I'm also here for this question
I figured out the answer. It's because pi is the angle between 2pi/3 and 4pi/3
Why is calc so hard 😩😩
bruh you're my nigga. I understand this now
how much does this man know
god bless you
who else has an exam tmao?
Bruh how the heck did that become pi+3 sqrt 3, 23:35
man the r 2 is looks like the sqrt(2) hahaah that is funny.
i think in 23:59 the answer is wrong. ans=8.88126
Can someone please tell me what I should do if I'll get a negative area?
Is it okay or not?
I did almost everything already but I can't seem to get my area to be positive... My brain's already dying
Does flipping the bounds work?
😋Sweet
This shi makes no sense
Shi got me tweaking
who the hell puts a one minute add on the video......
Only people who deserve it
bro you only show easy simple examples
Zarif Mahfuz this is a a channel for learning not crazy problems. If you want crazy problems go to flammable math or blackpenredpen or dr peyam.
Thanks