Question

Find the volume of the solid of revolution formed by rotating about the x--axis the region bounded by the given curves.

Answer 1

Answer:

$V = \pi [\frac{280}{3} -\frac{64}{3}]= 72\pi$

Step-by-step explanation:

For this case we are interested on the region shaded on the figure attached.

And we can find the volume with the method of rings.

The area on this case is given by:

$A(x) = \pi [f(x)]^2 = \pi r^2 = \pi [x-4]^2 = \pi (x^2 -8x +16)$

And the volume is given by the following formula:

$V= \int_{a}^b A(x) dx$

For our case our limits are x=4 and x=10 so we have this:

$V = \pi \int_{4}^{10} x^2 -8x +16 dx$

And if we do the integral we got this:

$V= \pi [\frac{x^3}{3} - 4x^2 +16 x]\Big|_4^{10}$

And after evaluate we got this:

$V=\pi [(\frac{1000}{3} -400 +160)-(\frac{64}{3} -64 +64)]$

$V = \pi [\frac{280}{3} -\frac{64}{3}]= 72\pi$