Using the shell method, calculate the value of the solid formed by$ y=(x+12)^ {1/2}$, $y=x$, and $y=0$ rotated about the$ x$-axis.

Question

So I am trying to figure out this question as I can't figure out how to get $h(y)$ for this question. I can find the answer online but I am having difficulty understanding why it is the answer which is $h(y) = (y+12-y^2).$ Any help is appreciated. Thank you in advance!

Answer 1

Draw a picture, something like. Then you can see that the cylindrical shells of radius $y$ start from the curve $y=\sqrt{x+12}$ and end up at the curve $y=x$. But the cylinder axis is along $x$ direction, so you need to extract $x$ from the above equations. So, to get the lower limit $$y^2=x+12$$ or $$x=y^2-12$$The upper limit is at $$x=y$$ The difference is the height of the cylinder: $$h(y)=y-(y^2-12)=y+12-y^2$$