Find the volume of the solid that results when the region enclosed by y=....?

Question

Find the volume of the solid that results when the region enclosed by y=sqrt(x), y=0, and x=9 is revolved about the line x=9. Use "pi" in the answer.

Thanks for the help!

Answer 1

Using the "shell method", the "height" of the shell is just f(x) = √x, and the "radius" is (9 - x):

⌠9
⌡ 2π(9-x)(√x)dx
0

.....⌠9
2π⌡ (9√x - x√x)dx
..... 0

2π[9(2/3)x^(3/2) - (2/5)x^(5/2)] from 0 to 9
2π[6x^(3/2) - 2/5x^(5/2)] from 0 to 9

Plugging in 0 just gives you 0, so:

2π[6(9)^(3/2) - 2/5(9)^(5/2)]
2π[6(27) - 2/5(243)]
2π[162 - 486/5]
2π/5[810 - 486]
648π/5 = 129.6π

If I didn't screw up.