Let R be the region bounded by the curve $\, y=\sqrt{x}\,$ and the lines $\, x=0\,$ and $\,y=5. $
What is the volume of the solid generated when R is rotated about the line $\, x=-4 \;$ ?
Okay, so what's the radius? $y = \sqrt{x}$?
And I need to subtract from it the "hole"/empty area… I tried $\sqrt{x} -4$ but it didn't work, any clues?
Also, the interval for integration is going to be $0$ to $25$, right?
Best Answer
It is easier to view the radius as a function of $y$ since you are rotating about a vertical line. This way, you may take the $x$ displacement from the origin, which will vary with $y$, and add 4 to obtain the radius.
Done correctly, you should get something like $r(y)=y^2+4$ and the integral $\pi \cdot \int_0^5{r^2(y)\;dy}$
Doing the integration and subtracting the cylinder that is indeed a "hole" like you pointed out should give you the result.