I'm having a little bit of trouble understanding how to determine the bounds and manipulate the equations of a complex solid of rotation problem. Here's a prime example of where I struggle:
Taken from a Calc 1 Textbook:
Determine the volume of a solid of rotation bounded by the curves $x=y^2$ and $x=4$ about the line $x=6$.
My attempt:
I determined it would be best to use the washer method because there is a "gap" between $x=4$ and $x=6$, and I assumed this gap would be the inner radius. As for the outer radius, I tried to use a piecewise function because my textbook said it is possible to use them when there is a "gap" like that (although I am definitely open to other ideas.)
$$ \begin{cases} x=y^2 & \text{if $x\in[0, 4]$} \\ 0 & \text{if $x\in[4,6]$} \end{cases}$$
From this, I set up the integrals.
$$\pi(\int_0^2(6-y^2)^2dy+\int_2^\sqrt{3}(6-0)^2dy)-\pi\int_0^2(6-y^2)^2dy$$
And I got: $$\pi(35\sqrt{3}-72)$$
Which is obviously wrong. Here's the right answer:
$$\frac{384\pi}{5}$$
As we can all see, I became very confused with this problem and mixed a few different methods together in a way that made no sense. Here's what I'm confused about exactly:
- Determining the bounds of the integrals in the washer method (when do I change them, when do I keep them the same)
- How to account for the "gap" that sometimes occurs (in my case, the gap between $x=4$ and $x=6$)
- When to use washer, disc, or shell
- When I use washer, how do I determine what is the outer radius and the inner radius (I understand it when I have to polynomials as the bounds, but I get confused in an example like this where it's just one polynomial and one vertical\horizontal line)
Thanks for your help.
You can use either method, as I will illustrate. Refer to the picture below.
First, the cylindrical shell method: this corresponds to the green shaded segment in the figure above. It is important to note that the $x$-coordinate ranges from $x = 0$ to $x = 4$, and on that interval, the height of the shell is $h(x) = y = 2\sqrt{x}$ (we count both the volume above and below the $x$-axis since we are not told that the volume is bounded by $y = 0$). The radius of the shell at $x$ is not $x$ itself, but $r(x) = 6-x$. So for example, when $x = 1$, the shell's height is $h(1) = 2$, and the radius is $r(1) = 5$.
So, for a differential thickness of shell $dx$, the differential volume is $$dV = 2\pi r(x) h(x) \, dx = 2\pi (6-x) 2\sqrt{x} \, dx.$$ We integrate this expression over $x \in [0,4]$ to obtain the total volume: $$V = \int_{x=0}^4 dV = 4 \pi \int_{x=0}^4 6 x^{1/2} - x^{3/2} \, dx = 4\pi \left[4x^{3/2} - \frac{2x^{5/2}}{5} \right]_{x=0}^4 = 4 \pi \left(32 - \frac{2}{5} 4^{5/2} \right) = \frac{384\pi}{5}$$ as claimed.
Now, we do the same calculation using washers. This corresponds to the red shaded segment in the picture above. In this case, the variable of integration is $y$, not $x$. The maximum value of $y$ occurs when $x = 4$, so $y = 2$, and by symmetry, $y = -2$ is the minimum value. So we can either integrate on $y \in [-2,2]$ or double the volume we get from integrating on $y \in [0,2]$. Let's do the latter. Then for a given $y \in [0,2]$, the inner radius of the washer is always $r_1(y) = 2$, and the outer radius is simply $r_2(y) = 6-y^2$. This is because the volume of revolution comprises a parabolic arc revolved around $x = 6$, which has been "drilled" through by a cylinder of radius $2$ about the same axis. Note that as a check, we can confirm that when $y = 0$, the outer radius should be $6$, corresponding to the distance from the axis of revolution $x = 6$ to $x = 0$ where the solid touches the $y$-axis. Now for a given $y \in [0,2]$, the differential thickness of such a washer is $dy$, and the differential volume is $$dV = \pi (r_2(y)^2 - r_1(y)^2) \, dy = \pi ((6-y^2)^2 - 2^2) \, dy.$$ The full volume is therefore $$V = 2 \int_{y=0}^2 \pi ((6-y^2)^2 - 4) \, dy = 2\pi \int_{y=0}^2 32 - 12y^2 + y^4 \, dy = 2\pi \left[ 32 y - 4y^3 + \frac{y^5}{5} \right]_{y=0}^2 = 2\pi \left(64 - 32 + \frac{32}{5} \right) = \frac{384\pi}{5}.$$