MHB B's question at Yahoo Answers involving a solid of revolution

[MarkFL]

Here is the question:

Find the volume V of the solid obtained by rotating the region bounded by x=16-(y-3)^2, x+y=7 about the x-axis?

Here is a link to the original question:

Find the volume V of the solid obtained by rotating the region bounded by x=16-(y-3)^2, x+y=7 about the x-axis? - Yahoo! Answers

I have posted a link there to this topic so the OP can find my response.

First, let's take a look at the region to be revolved:

View attachment 547

As we can see, it will be cleaner to consider the shell method.

The volume of an arbitrary shell is:

$\displaystyle dV=2\pi rh\,dy$

$\displaystyle r=t$

$\displaystyle h=\left(16-(y-3)^2 \right)-(7-y)=7y-y^2$

and so we have:

$\displaystyle dV=2\pi y(7y-y^2)\,dy=2\pi(7y^2-y^3)\,dy$

Now, to determine the limits of integration, we want to find the roots of:

$\displaystyle h(y)=0$

$\displaystyle 7y-y^2=0$

$\displaystyle y(7-y)=0$

$\displaystyle y=0,\,7$

And so, summing the shells by integration, we find:

$\displaystyle V=2\pi\int_0^7(7y^2-y^3)\,dy=2\pi\left[\frac{7}{3}y^3-\frac{1}{4}y^4 \right]_0^7=2\pi\cdot7^4\left(\frac{1}{3}-\frac{1}{4} \right)=\frac{7^4\pi}{6}=\frac{2401\pi}{6}$

 

[Prove It]

I tend to prefer the method of washers. The volume can be calculated as

$\displaystyle \begin{align*} V &= \int_0^7{\pi \left[ 16 - (y - 3)^2 \right]^2 \, dy} - \int_0^7 {\pi \left( 7 - y \right)^2 \,dy} \\ &= \pi \int_0^7{ y^4 - 12y^3 + 22y^2 + 84y + 49 \,dy } - \pi \int_0^7{ 49 - 14y + y^2 \, dy } \\ &= \pi \int_0^7{ y^4 - 12y^3 + 21y^2 + 98y \,dy } \\ &= \pi \left[ \frac{y^5}{5} - 3y^4 + 7y^3 + 49y^2 \right]_0^7 \\ &= \pi \left[ \left( \frac{16\, 087}{5} - 7203 + 2401 + 2401 \right) - \left( \frac{0}{5} - 0 + 0 + 0 \right) \right] \\ &= \frac{4802\pi}{5} \end{align*}$

Edit: It appears I rotated this around the y-axis instead of the x. Hope it was helpful anyway :)