# Region bounded by the curve $\, y=\sqrt{x}\,$ and the lines $\, x=0\,$ and $\,y=5.$ What's the volume when rotated about x=-4?

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?

• Yes, your interval is correct. If I read the problem correctly, your radius should be $29$, but the first $4$ units (the hole you mentioned) closest to the centre are not part of the solid. – Lanier Freeman Sep 10 '15 at 2:07
• @L.A.F.2. so how would do I subtract the other radius? I can't just say R = 29, and r = 4? then 29-4, how I'd integrate it... doesn't seem right. – Jack Sep 10 '15 at 2:14
• If you are doing it by slicing (washers) you will be integrating with respect to $y$. The outer radius at height $y$ is $4+x$, that is, $4+y^2$, and the inner radius is $4$. Or else you can do it by cylindrical shells, in which case you will be integrating with respect to $x$, radius $4+x$, "height" of the cylindrical shell $5-\sqrt{x}$. – André Nicolas Sep 10 '15 at 2:41
• I'd recommend using the Shell Method. Edit: Andre Nicolas' post explains how you can go about doing this if you don't already know. – Lanier Freeman Sep 10 '15 at 2:41
• I don't see any subtraction there... I still don't know what my formula looks like. – Jack Sep 11 '15 at 1:26

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 $\pi \cdot \int_0^5{r^2(y)\;dy}$