You realize that by trying to combine steps, and then finding errors, is a great way to emphasize the necessity of tedious, stepwise derivations. No doubt this is purely coincidental, and not part of your plan for this lesson. Nice
You bring up a good point. My emphasis usually centers around how do we present this so that that student can understand what we are doing and how can I present it in a way that is common to many of these types of problems, so that they can learn the specific methodology of solving that type of problem.
Besides the end substitution for pi, that instead of pi/2- he squared and it became pi/4, the brilliant part was substituting for y=sin (theta) and then using the trigonometric identity.
I used a dx instead of a dy and still got the same answer. I wouldn't have thought to try that (less conventional) method if I didn't watch the previous video on this series so thank you!!
Dear Sir, Many Thanks for this wonderful video. It was as great as always. I am just wondering what if the circle equation was x^2+y^2= 9...Then the limits of integration would have been 0 and 3. In that case could we still use trig substitution? If not, would it be still possible to calculate the volume?
The result may be wrong..! Think a full toroid and try to find the whole volume of it. R=1, so the area is (Pi) and multiply it by the perimeter to get the volume. The result will be (Pi)*(2*Pi*2)=4*(Pi)^2. In this example, there is a quarter toroid which volume is being calculated for. So, the requested volume should be (Pi)^2.
Hello Professor, do you think it was also possible to calculte dV=πR(h)dx --> dV=π(2-x)( √(i-x^2))dx , using the same method of the previous lecture? (i tried but i came up with a different result =π2/3 ) Thank you, Best Regards
I tried it and got the same dv as you did but I got the correct answer = pi((3pi-2)/6) --approx. 3.887. You must've made a mistake in the integrals PS: It is the root of ( *1* -x^2) not i
That is a trick that just pops into one's head. It is a trick that we learn and it comes in handy when dealing with integration that involves radicals.
You realize that by trying to combine steps, and then finding errors, is a great way to emphasize the necessity of tedious, stepwise derivations. No doubt this is purely coincidental, and not part of your plan for this lesson. Nice
You bring up a good point. My emphasis usually centers around how do we present this so that that student can understand what we are doing and how can I present it in a way that is common to many of these types of problems, so that they can learn the specific methodology of solving that type of problem.
Besides the end substitution for pi, that instead of pi/2- he squared and it became pi/4, the brilliant part was substituting for y=sin (theta) and then using the trigonometric identity.
I used a dx instead of a dy and still got the same answer. I wouldn't have thought to try that (less conventional) method if I didn't watch the previous video on this series so thank you!!
There are often multiple ways in which these types of problems can be solved. Thanks for the insight.
Dear Sir,
Many Thanks for this wonderful video. It was as great as always.
I am just wondering what if the circle equation was x^2+y^2= 9...Then the limits of integration would have been 0 and 3. In that case could we still use trig substitution? If not, would it be still possible to calculate the volume?
It still can be done, but it will become more difficult to manipulate the conversion.
if it's a circle and bounded by x axis and y axis. isn't all 4 quadrants are bound inside?. confused. 😐
The result may be wrong..! Think a full toroid and try to find the whole volume of it. R=1, so the area is (Pi) and multiply it by the perimeter to get the volume. The result will be (Pi)*(2*Pi*2)=4*(Pi)^2. In this example, there is a quarter toroid which volume is being calculated for. So, the requested volume should be (Pi)^2.
dear sir there is a mistake at 11:35 of the video ; the cos(theta) should be used instead of [1/2(1+cos(theta))]
sir u were mistaken at the last part. should have been phi^2/2 instead of phi^2/4 because it is phi times phi/2
You are correct.
Awesome channel
Awesome content
Hello Professor, do you think it was also possible to calculte dV=πR(h)dx --> dV=π(2-x)( √(i-x^2))dx , using the same method of the previous lecture? (i tried but i came up with a different result =π2/3 ) Thank you, Best Regards
I tried it and got the same dv as you did but I got the correct answer = pi((3pi-2)/6) --approx. 3.887. You must've made a mistake in the integrals
PS: It is the root of ( *1* -x^2) not i
Sir, Just one question. How do you know how to 'slice' the shape. I mean, how do we know how to do it. Any tips or tricks?
watch previous lessons
This was a tough problem. I didnt' know you could just substitute y for a trig function.
Yes, that can simplify such problems.
I find drawing a right triangle helps. In this case 1 would be the hypotenuse, and y would be either the adjacent or opposite side of the triangle.
thanks
You're welcome!
let y = sin(θ) => super awesome
That is a trick that just pops into one's head. It is a trick that we learn and it comes in handy when dealing with integration that involves radicals.
Super!!!! Thank you.
Answer is wrong. Answer you've been [pi(pi/2 - 1/3)] not [pi(pi/4 - 1/3)]
i wasted 20 min 💀💀