# Thread: Volume of the solid of revolution

1. ## Volume of the solid of revolution

There are a region bounded by the curve y=cos^2(x)sin^4(x) and the x-axis , where pi<x<2pi. Find the volumn of the solid of revolution when the region is revolved about the y-axis

2. ## Re: Volume of the solid of revolution

Hi Lau, what have you tried already? It's not clear if you are stuck on what method to use to solve this or if the integration is too difficult.

3. ## Re: Volume of the solid of revolution

Hey Lau, I solved this problem using the shell method.

It is best to start with a rough sketch of the function on the given interval (Will give a M shape), and once you do you will realize that we should split the region again this time considering: pi <x<3pi/2 and 3pi/2<x<2pi. This is because from x=pi to x=3pi/2 , the function "starts" then "ends". Also, this will help us in the calculations as the function "repeats" its self on 3pi/2 < x < 2pi.

Using the shell method, I found the height of a shell to be: $\displaystyle h=f(x)= \sin^4x \cos^2x$ and a radius about the y axis to be $\displaystyle r=x$.

Then the volume of the first cycle of the periodic function is:

$\displaystyle V= \int_{\pi}^{3 \pi/2} 2 \pi r \cdot h \, dx$
$\displaystyle V= \int_{\pi}^{3 \pi/2} 2 \pi (x) \cdot (\sin^4x \cos^2x) \, dx = \frac{\pi}{36} + \frac{5 \pi}{64}$

So finally, the total volume is just $\displaystyle 2V = \frac{\pi}{18} + \frac{5 \pi}{32} \approx 5.02$

4. ## Re: Volume of the solid of revolution

Originally Posted by sakonpure6
Hey Lau, I solved this problem using the shell method.
It is best to start with a rough sketch of the function on the given interval (Will give a M shape), and once you do you will realize that we should split the region again this time considering: pi <x<3pi/2 and 3pi/2<x<2pi. This is because from x=pi to x=3pi/2 , the function "starts" then "ends". Also, this will help us in the calculations as the function "repeats" its self on 3pi/2 < x < 2pi.
Using the shell method, I found the height of a shell to be: $\displaystyle h=f(x)= \sin^4x \cos^2x$ and a radius about the y axis to be $\displaystyle r=x$.
Then the volume of the first cycle of the periodic function is:

$\displaystyle V= \int_{\pi}^{3 \pi/2} 2 \pi r \cdot h \, dx$
$\displaystyle V= \int_{\pi}^{3 \pi/2} 2 \pi (x) \cdot (\sin^4x \cos^2x) \, dx = \frac{\pi}{36} + \frac{5 \pi}{64}$

So finally, the total volume is just $\displaystyle \color{red}{2V = \frac{\pi}{18} + \frac{5 \pi}{32} \approx 5.02}$
@sakonpure6, you have a conceptual mistake.

$\displaystyle 3.30405 \approx \int_{1.5\pi }^{2\pi } {\left( {2\pi x} \right)\left( {{{\sin }^4}(x){{\cos }^2}(x)} \right)dx} \ne \int_\pi ^{1.5\pi } {\left( {2\pi x} \right)\left( {{{\sin }^4}(x){{\cos }^2}(x)} \right)dx \approx 2.0963}$

Look at the graph again.

5. ## Re: Volume of the solid of revolution

Originally Posted by Plato
@sakonpure6, you have a conceptual mistake.

$\displaystyle 3.30405 \approx \int_{1.5\pi }^{2\pi } {\left( {2\pi x} \right)\left( {{{\sin }^4}(x){{\cos }^2}(x)} \right)dx} \ne \int_\pi ^{1.5\pi } {\left( {2\pi x} \right)\left( {{{\sin }^4}(x){{\cos }^2}(x)} \right)dx \approx 2.0963}$

Look at the graph again.
EDIT a Typo: 2.50963 not 2.0963

6. ## Re: Volume of the solid of revolution

Oh thanks for pointing that out Plato! So, Lau the correct answer is: 3.304 +2.096 = 5.40

Edit: Plato, did something in the equation give away that the function was not periodic - had different amplitudes? Or did you plot to verify?

7. ## Re: Volume of the solid of revolution

It is best to start with a rough sketch of the function on the given interval (Will give a M shape), and once you do you will realize that we should split the region again this time considering: pi <x<3pi/2 and 3pi/2<x<2pi. This is because from x=pi to x=3pi/2 , the function "starts" then "ends". Also, this will help us in the calculations as the function "repeats" its self on 3pi/2 < x < 2pi.

8. ## Re: Volume of the solid of revolution

Originally Posted by dungtuyet
It is best to start with a rough sketch of the function on the given interval (Will give a M shape), and once you do you will realize that we should split the region again this time considering: pi <x<3pi/2 and 3pi/2<x<2pi. This is because from x=pi to x=3pi/2 , the function "starts" then "ends". Also, this will help us in the calculations as the function "repeats" its self on 3pi/2 < x < 2pi.
This is wrong. In fact, it completely misses the point of this whole thread!
@dungtuyet, did you even read the entire thread? If you did, what part did you not get?

One more time. Here is a standard example.
What is the volume of the graph of $y=|x-3|,~ 2\le x\le 4$ about the $y \text{-axis}~?$

Please study each of these: from $x=2\to 3$; then from $x=3\to 4$

Note that those two volumes are not equal even though the graph is symmetric about $x=3$. The radii are different?

Add those two values and compare the result with this.