Finding volume of a function rotating around the x-axis

**softballchick** · March 6th 2011, 10:22 AM

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...

**NOX Andrew** · March 6th 2011, 12:05 PM

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

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

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

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

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

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

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