Volume of Revolution - Cylindrical Shell

**VectorRun** · January 28th 2011, 05:05 PM

So I'm suppose to find the volume of revolution of an area enclosed by $y=2-x^{2}$ and $y=x^2$ rotated about x=1.

So I'm using the cylindrical shell method $\int_a^b 2\pi rhdx$ and I'm integrating with respect to x; finding the integration points for the limits of integration, I get b=1 or a=-1.

Since the axis is x=1, then r=1-x. The height, $h=(2-x^{2})-(x^2)=2-2x^2$

Therefore, the definite integral is (factoring out the 2pi from the start and 2 from the height): $4\pi \int_{-1}^{1} ((1-x)(1-x^2))dx$ .

Since it's symmetrical I can change the limits to 0 and 1, multiplying the integral by a factor of 2. Then expanding yields: $8\pi \int_{0}^{1} (1-x-x^2+x^3)dx$

By FTC: $=8\pi (x-\frac {x^2}{2}-\frac {x^3}{3}+\frac {x^4}{4})$ evaluated from 0 to 1.

That yields $8\pi (1-\frac {1}{2}-\frac {1}{3}+\frac {1}{4})$ and simplifying gets: $\frac {10\pi}{3}$

However, I'm apparently wrong as the answer key says that it should be $\frac {16\pi}{3}$ . I can't see my mistake, anyone see where I went wrong?

**TheEmptySet** · January 28th 2011, 05:55 PM

Originally Posted by VectorRun

So I'm suppose to find the volume of revolution of an area enclosed by $y=2-x^{2}$ and $y=x^2$ rotated about x=1.

So I'm using the cylindrical shell method $\int_a^b 2\pi rhdx$ and I'm integrating with respect to x; finding the integration points for the limits of integration, I get b=1 or a=-1.

Since the axis is x=1, then r=1-x. The height, $h=(2-x^{2})-(x^2)=2-2x^2$

Therefore, the definite integral is (factoring out the 2pi from the start and 2 from the height): $4\pi \int_{-1}^{1} ((1-x)(1-x^2))dx$ .

Since it's symmetrical I can change the limits to 0 and 1, multiplying the integral by a factor of 2. Then expanding yields: $8\pi \int_{0}^{1} (1-x-x^2+x^3)dx$

By FTC: $=8\pi (x-\frac {x^2}{2}-\frac {x^3}{3}+\frac {x^4}{4})$ evaluated from 0 to 1.

That yields $8\pi (1-\frac {1}{2}-\frac {1}{3}+\frac {1}{4})$ and simplifying gets: $\frac {10\pi}{3}$

However, I'm apparently wrong as the answer key says that it should be $\frac {16\pi}{3}$ . I can't see my mistake, anyone see where I went wrong?

Remember that

$\int_{-1}^{1}x^ndx=0$ if $n$ is odd and

$\int_{-1}^{1}x^ndx=2\int_{0}^{1}$ if $n$ is even

This gives

$\displaystyle 8\pi \left(1-\frac{1}{3}\right)=\frac{16}{3}\pi$

**HallsofIvy** · January 28th 2011, 06:10 PM

The cylindrical shell method divides the solid into cylinders having height $y_2- y_1$ and radius equal to the distance from the axis of rotation. Here, $y_2- y_1= (2- x^2)- (x^2)= 2- 2x^2$ and the radius is x-(-1)= x+ 1. The surface area of such a cylinder is $\pi r^2 h= \pi(2- 2x^2)(x+1)$ and, taking "dx" to be the thickness of the cylinder, the volume is the integral $\pi\int_{-1}^1 (2- 2x^2)(x+ 1)dx$ .

Yes, $y_2- y_1= 2- 2x^2$ is symmetrical about x= 0 but the radius, x+ 1, is NOT.

**VectorRun** · January 28th 2011, 06:29 PM

Right! I can't believe I overlooked that. Alrighty, I figured out the rest from there. Splitting up the integrals and then applying the symmetry rules. I'm definitely getting $\frac{16}{3}\pi$ now. Thanks a lot!

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