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$x = 0, x = 9 - y^2$ rotated about $x = -1$

I'm having a lot of trouble deciding whether to use the disc method or the shell method. Intuitively, it makes sense that the shell method would be simpler when you are rotating horizontally, like around the y-axis or x = -1.

I know that the shell method is:

$[circumference][height][width]$, where $C = 2\pi r$, $h = f(x)$, and $w = \Delta x$

I must find the radius of the solid about the line $x = -1$.

So, I see that the radius must be $y - (-1) \rightarrow r = y + 1$

Therefore, $C = 2\pi (y + 1)$, right?

The height varies with $f(y) = 9 - y^2$.

Here is where things get really muddled for me.

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up vote 1 down vote accepted

If it's been said once, it's been said seventeen times, and worth saying again: draw the picture:

enter image description here

Using the washer method, you would integrate with respect to $y$ from $y=-3$ to $y=3$ and

  • the outer radius of the washer in terms of $y$ is $\color{darkblue}1+\color{darkgreen}{d_y}=1+(9-y^2)$
  • the inner radius of the washer in terms of $y$ is $\color{darkblue}1$.

So the integral would be $\pi\int_{-3}^3 (10-y^2)^2-1^2\,dy$.

Using the shell method, you would integrate with respect to $x$ from $x=0$ to $x=9$ and:

  • The height of the shell in terms of $x$ is $2\cdot\color{gray}{h_x}=2\sqrt{9-x}$.
  • The radius of the shell in terms of $x$ is $\color{darkblue}1+\color{darkgreen}{d_y}=1+x$.

    So the integral would be $2\pi\int_0^9 (1+x)\cdot 2\sqrt{9-x}\,dx $.

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How do you get the graph like that? – Jonathan Dewein May 3 '12 at 19:01
@JonathanDewein Using JSXgraph – David Mitra May 3 '12 at 19:06

The integral in question can be solved by moving to cylindrical coordinates. In these coordinates: $$9 - h^2 + 1 > r > 1 $$ $$ 3 > h > -3$$ $$ 2\pi > \phi > 0$$ Your integral is therefore: $$\int_0^{2\pi}\int_{-3}^{3}\int_1^{10-h^2}r\sin(\phi)drdhd\phi$$

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Er, sorry man, but I'm only in calculus I. I have no idea how to do that. – Jonathan Dewein May 3 '12 at 17:10
@JonathanDewein - first preform the integration over $r$. This will give you a function of $h$. Then do the integral over $h$, and finally the integral over $phi$. – nbubis May 3 '12 at 17:44

The curve is a backward opening parabola. I assume you have drawn it.

For shells, we are taking a thin vertical strip through our planar region, from $x$ to $x+dx$, and rotating the strip about $x=-1$. So the radius of the circle is $x-(-1)=x+1$, and not your $y+1$. As to the height, it is "$2y$." You can use $x=9-y^2$ to find that $y=\sqrt{9-x}$. So we will be integrating $(2\pi(1+x))(2\sqrt{9-x})$ from $x=0$ to $x=9$. To find the integral, let $9-x=u$.

Horizontal slicing may be a little easier, though we need to deal with the hole. We will be integrating with respect to $y$. It is better to find the volume of the top half and double.

The "washer" at height $y$ has outer radius $x+1$, and inner radius $1$, so area $\pi((1+x)^2-1^2)$. We will be integrating with respect to $y$, so use the fact that $1+x=10-y^2$. Integrate from $y=0$ to $y=3$, then double.

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