# Math Help - Volume of a revolving solid...

1. ## Volume of a revolving solid...

I don't feel very confident with this concept.

Suppose R is the region bounded by y = x2, x = 2, and y = 0

1) Find the volume of a solid that is generated by revolving R about the x-axis. - I believe I did this one correctly, but I want to be sure. I got 32pi/5.
2) (same) ... revolving R about the y-axis.
3) ... revolving R about y = -3.
4) ... revolving R about x = 3.

For 3 and 4, I've tried setting up the integrals a few different ways, but I don't think any of them are right.

If someone could help me set these up or give me some tips & verify that the first two are correct, I'd be eternally grateful.

2. ## Re: Volume of a revolving solid...

1) \displaystyle \begin{align*} V = \pi \int_0^2{ \left( x^2 \right) ^2 \, dx } \end{align*}

2) \displaystyle \begin{align*} V = \pi \int_0^4{ 2^2 - \left( \sqrt{y} \right) ^2 \, dy } \end{align*}

3) \displaystyle \begin{align*} V = \pi \int_0^2{ \left( x^2 + 3 \right) ^2 \, dx } \end{align*}

4) \displaystyle \begin{align*} V = \pi \int_0^4{ \left( 3 - \sqrt{y} \right) ^2 - 1^2 \, dy} \end{align*}

3. ## Re: Volume of a revolving solid...

Lets look at this step by step. Drawing a line from the given curve to the axis of rotation gives a radius of the circle the point on the curve follows as the curve is rotated around the axis. The area of the disk formed is $\pi r^2$. Taking the "thickness" of that line to be the differential "du" (u is perpendicular to the radius so along the axis of rotation) the volume is $\pi r^2du$. The volume of the entire figure is $\int \pi r^2 du$.

Suppose R is the region bounded by y = x2, x = 2, and y = 0

1) Find the volume of a solid that is generated by revolving R about the x-axis. - I believe I did this one correctly, but I want to be sure. I got 32pi/5.
The distance from the point (x, y) to (x, 0) (the x-axis) is $y= x^2$ so $r= x^2$, $r= (x^2)^2= x^4$ and the integral is $\pi \int_0^2 x^4 dx$. Yes, that is $\frac{32}{5}\pu$.

Since the region does not extend all the way to the y-axis, this really a "washer". But you can do it as two disks. Calculate the volume to the outer boundary, x= 2, calculate the volume to the inner boundary, $x= \sqrt{y}$, and subtract.
The radius of the "outer boundary" is the constant, 2, so rotation forms a cylinder with radius 2 and height $2^2= 4$ so volume $\pi r^2h= \pi (4)(4)= 16\pi$. The radius of the "inner boundary" is $\sqrt{y}$ so each disk has area $\pi r^2= \pi y$. Taking the "thickness" to be $dy$, the volume of the outer portion is $\pi \int_0^4 ydy= 8\pi$. The difference is $16\pi- 8\pi= 8\pi$, as you say.

3) ... revolving R about y = -3.
The distance from $y= x^2$ to y= -3 is $r= x^2- (-3)= x^2+ 3$ so the area of each "disk" is $\pi(x^2+ 3)^2$ and, taking "dx" as the thickness the volume of each disk is $\pi(x^2+ 3)^2 dx$. The total volume is $\pi \int_0^2 (x^2+ 3)^2 dx$.

4) ... revolving R about x = 3.
The distance from (x, y) to (3, y) is $r= 3- x= 3- \sqrt{y}$ (x= 3 is to the right of the region). The area of a disk is $\pi r^2= \pi(3- \sqrt{y})^2= \pi(y- 6\sqrt{y}+ 9)$. Taking the thickness of a disk to be dy, the volume of each disk is $\pi (y- 6\sqrt{y}+ 9)dy$ and the total volume is $\pi\int_0^4(y- 6y^{1/2}+ 9)dy$.

4. ## Re: Volume of a revolving solid...

Thanks so much to both of you - ProveIt for the quick response, and HallsofIvy for going in-depth.