Find the volume of the solid that results when the region is revolved...?

Question

Find the volume of the solid that results when the region enclosed by the given curves is revolved about the x-axis. (Use "pi") Do not round any coefficients.

y=sqrt(100-x^2), y=6

Thanks everbody X^D

Answer 1

First and most important is to determine the bounds for the tripple integral.
We will be using folowing notations:
r instead of y (because it reflects the radius of the rotational solid)
z instead of x (because it reflects the height of the rotational solid)
§{low, hi} as the simbol for integral.

Given the above notations volume of each rotational solid can be represented with the following formula:

(1) V = 2Pi * §§r dr dz

So we need to establish the boundaries for r and z
If we sketch the region rotated we can plainly see that

(2) z goes from -8 to 8

And since the region boundaries are given by the two curves, the skech shows that

(3) r goes from 6 to sqrt(100 - z^2)

Given that we have that

(4) V = 2Pi §{-8, 8} §{6, sqrt(100 - z^2)} r dr dz

Solving the integral by r is rather easy. Doing that we get

(5) V = 2Pi §{-8, 8} r^2/2 |{6, sqrt(100 - zˇ2)} dz

which gives

(6) V = Pi §{-8, 8} (100 - z^2 - 36) dz = Pi §{-8, 8} (64 - z^2) dz

Solving the other integral (which is also an easy one) we get

(7) V = Pi(64 §{-8, 8} dz - §{-8, 8} z^2 dz) = Pi(1024 - z^3/3|{-8, 8})

Which finally yields the result

(8) V = Pi(1024 - 1024/3) = 2048 Pi/3