Find the volume formed by rotating the enclosed region

Question

Find the volume formed by rotating the enclosed region $y=4\sqrt{x}$ and $y=x$ about $x=17$

I have tried plugging everything into the formula but I can't seem to get the right answer.

How do I solve this?

Answer 1

Hint:

Inner Radius = $17-y$

Outer Radius = $(17-\frac{y^2}{16})$

Points of intersection = 0,16

Volume Enclosed by the region $=\int_0^{16} \pi\left((Outer Radius)^2 - (Inner Radius)^2\right)dy$

$$V = =\int_0^{16} \pi\left((17-\frac{y^2}{16})^2 - (17-y)^2\right)dy$$ If you evaluate this you get the volume

$$V = \frac{13568\pi}{15}$$

Answer 2

As a hint, you need to get the area between the two functions on the interval x=0 to x=17:

${y=4\sqrt{x}}$ and y=x.

To do this, you'll need to use definite integrals:

${\int_0^{16} \! 4\sqrt{x} \, \mathrm{d}x - \int_0^{16} \!x \, \mathrm{d}x}$

Now that we have the area between these functions, picture them rotating around x=17. Remember that volume is a sum of areas. Can you think of an integral involving angles, that would sum up this area into a volume? Specifically involving theta rotating from ${0}$ to ${2\pi}$.

*Think "shell method or cylinder method?"