[Math] Find the volume of the solid by rotating the given curves about the specified axis.

calculusintegration

$x = 0, x = 9 – y^2$ 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.

Best Answer

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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