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The volume of the solid from the region bounded by $x=9-y^2$, $y=x-7$, $x=0$ about $y=3$ using cylindrical shells.

I've tried creating two separate regions:

$V_1=2\pi(3-y)(9-y^2)dy$ from 3 to 1

and

$V_2=2\pi(3-y)(y+7)dy$ from 1 to 0

but the answer this gives isn't correct. I can't seem to find any errors in my calculations. Am I devising the solution incorrectly?

  • 0
    Ian, you can accept answers that you find particularly helpful. (You can upvote as many answers as you like, by clicking on the arrow above the "vote count" on the left of the answer, and can accept one answer per question by clicking on the "greyed out" check-mark to the left of that answer.2013-01-15

2 Answers 2

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Here again is the graph of the things mentioned in the problem:

$\color{blue}{x=9-y^2}$, $\color{red}{x=y+7}$, $\color{green}{x=0}$, $y=3$ (dashed)

enter image description here

You're correct that the circumference of the shell at $y$ is given by $2\pi(3-y)$, but you should be integrating over all of $-3\leq y\leq 3$, with the height of the shell at $y$ given by $\begin{cases} 9-y^2&\text{ if }1\leq y\leq 3,\\ y+7 & \text{ if }-2\leq y\leq 1,\\ 9-y^2 & \text{ if }-3\leq y\leq -2.\end{cases}$ Thus, the volume is $V=\int_1^3 (3-y)(9-y^2)\,dy+\int_{-2}^1(3-y)(y+7)\,dy+\int_{-3}^2(3-y)(9-y^2)\,dy$ which you can check gives the right answer.

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Let's see if this picture and integral set up are enough to get you over the hump:

Mathematica graphics

The arrows might help you picture the heights if the various shells.


Here's the washer method I accidentally posted first, just in case it might be helpful to you to think about the problem a different way. The arrows denote the inner and outer radii of the washers.

Mathematica graphics