# Math Help - Finding volume of a function rotating around the x-axis

1. ## Finding volume of a function rotating around the x-axis

Find the volume of the solid formed by rotating the region inside the first quadrant enclosed by and about the -axis.

So I know it is pi[ outer^2-inner^2]

the boundaries are 0 to 2. and I know that y=8x is on top.

But I can't seem to get the right answer, I would get a negative number...

2. Does the integral look like $\displaystyle \pi \int_0^2 [(8x)^2 - (x^4)^2] \, dx$? If so, then the answer is $\dfrac{1024\pi}{9}$.

$\displaystyle \pi \int_0^2 [(8x)^2 - (x^4)^2] \, dx$

$\displaystyle = \pi \int_0^2 (64x^2 - x^8) \, dx$

$\displaystyle = \pi [\dfrac{64}{3}x^3 - \dfrac{1}{9}x^9]_0^2$

$\displaystyle = \pi (\dfrac{64}{3}2^3 - \dfrac{1}{9}2^9)$

$\dfrac{1024\pi}{9}$