# Volume of a solid of revolution bounded by

• April 11th 2011, 10:48 AM
bijosn
Volume of a solid of revolution bounded by
Determine the volum of a solid of revolution when the region bounded by $y = \sqrt{x}$ and the straight lines y = 2 and x = 0 are rotated around the line x = 4

$V = 2\pi \int_{0}^{4} x ( 2 - \sqrt{x}) dx$
$V = 2\pi ( x^2 - \frac{2x^\frac{5}{2}}{5} )\big|_0^4$

$= \frac{32\pi}{5}$
??
what am I doing wrong?
• April 11th 2011, 12:16 PM
Unknown008
When you rotate about the x-axis (y = 0), you integrate with respect to x.

When you rotate along the line x = 4, you integrate with respect to y.

And can you tell why you picked $x(2 - \sqrt{x})$?
• April 11th 2011, 12:21 PM
Prove It
Hint: Translate your function and all the bounds left 4 units. The region does not change, but the calculations are easier because you can rotate around the x axis.