Use the method of cylindrical shells to find the volume of the solid obtained by rotating the region bounded by the given curves about the x-axis.

$$x = 1+(y - 2)^2, \quad x = 2$$

this is confusing because there is no bounds on the left side of the graph. also I have the equation 2pi(int)(a-b)yf(y)dy and i know y is the radius but I'm not sure how to manipulate the formula to reflect different changes in the radius in general..


1 Answer 1


We need a picture. Note that the curve is a rightward-opening parabola, with axis $y=2$, and vertex at $(1,2)$. If you have a graphing calculator, it will help you to visualize. The free program Wolfram Alpha should also do a decent job of graphing.

We will be integrating with respect to $y$. By solving $1+(y-2)^2=2$. we find that the line $x=2$ meets the parabola at $y=1$ and $y=3$.

Take a thin horizontal strip at height $y$. This is at distance $y$ from the $x$-axis, so the radius of our shell is $y$.

The length of the thin horizontal strip is $2-x$, that is, $2-\left((y-2)^2+1\right)=1-(y-2)^2$. Thus our volume is $$\int_{y=1}^3 2\pi y (1-(y-2)^2)\,dy.$$

  • $\begingroup$ oh.. I thought the bottom half of the parabola was undefined when graphing y = √(x-1)+2 $\endgroup$
    – J L
    Commented Sep 16, 2013 at 4:47
  • $\begingroup$ i must have left out the + or - part when taking the square root $\endgroup$
    – J L
    Commented Sep 16, 2013 at 4:48
  • $\begingroup$ wait why did you put 2-y instead of y-2?? $\endgroup$
    – J L
    Commented Sep 16, 2013 at 5:01
  • $\begingroup$ No good reason, and since it may cause confusion I have changed it to $y-2$. Of course it doesn't matter, since we are squaring it. $\endgroup$ Commented Sep 16, 2013 at 5:07
  • $\begingroup$ ok good. ______ $\endgroup$
    – J L
    Commented Sep 16, 2013 at 5:08

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