integrationmultiple integralmultivariable-calculussolid of revolutionvolume
The problem goes like this
Rotate the indicated area around the given axis to calculate the volume of the solid of revolution $y=x^2+1$, $x=0$, $x=2$, $y=0$ , around the axis $ y = 5$.
My question is if the solution is given by the following integral (should I take the limits of integration from $0$ to $2$?)
$$2\pi\int_0^2 ((5-y)\sqrt{y-1})\,dy$$
Visualization is an important feature of getting the integral set up correctly for volumes of revolution. The shell method is more complicated for this problem because the shell widths vary as differences between two sets of different functions. However, I'll proceed anyway with that method.
A shell method rotated about the $y$ axis will have shell thicknesses of $dx$ so we need to express the integral in terms of $x$. Each shell will be of height $4 - \frac{x}{2}$ from $x=\frac{4}{3}$ to $x=8$ (for the first limit of integration), and height $3x-\frac{x}{2}$ from $x=0$ to $x=\frac{4}{3}$ (for the second limit of integration).
The basic formula is:
$$\int_a^b 2\pi(\text{shell radius})(\text{shell height}) dx$$
We have two separate integrals to add to get the total volume. $$\int_{\frac{4}{3}}^8 2\pi(x)(4 - \frac{x}{2}) dx + \int_{0}^{\frac{4}{3}} 2\pi(x)(3x-\frac{x}{2}) dx$$
Which simplifies to: $$2\pi \int_{\frac{4}{3}}^8 4x - \frac{x^2}{2} dx + 2\pi \int_{0}^{\frac{4}{3}} \frac{5x^2}{2}dx$$ $$2\pi\cdot ([2x^2 - \frac{x^3}{6}]_{\frac{4}{3}}^8 + [\frac{5x^3}{6}]_0^{\frac{4}{3}})$$ $$2\pi\cdot (128 - \frac{256}{3} - \frac{32}{9} + \frac{64}{162} + \frac{320}{162})$$ $$= 260.6358$$
I can see how it can be a bit confusing that the given region lies both above and below the $x$-axis, about which we're supposed to revolve it. The question does make sense, though. And if we imagine actually revolving it, we will see that — thanks to the symmetry of the region — the solid we get is the same as if we revolve the upper half (lying above the $x$-axis) only.
Next, if you decide to use this method (as opposed to the method of cylindrical shells, which I feel may be a little easier here), then you did make a mistake in how you set it up in your post. Look at the picture (where I still shaded the entire region, not just the upper half of it):
The second curve $y^2=2(x-3)$ does NOT extend all the way to the left to $x=0$. In other words, you do NOT have an inner radius of $r=\sqrt{x-3}$ between $x=0$ and $x=3$. So your setup with an inner and outer radius is only correct between $x=3$ and $x=6$, but NOT between $x=0$ and $x=3$. In that part, there's only one radius, since there's no hole in the middle.
Thus, to correct your solution, you need to split the integral into two parts: $$\text{Volume}=\int_0^3\text{something}\cdot dx+\int_3^6\text{something}\cdot dx.$$ What you did in your post goes into the second integral, but there's gonna be something different (and more simple) in the first one.
Best Answer
As you are integrating with respect to $y$, it should be $1 \leq y \leq 5$. But you are finding volume of region $2$ rotated around $y = 5$. You can do that but then you need to subtract it from the volume of the cylinder of radius $5$ to find volume of region $1$ that the question asks.
An easier way to do this is to directly find the volume of region $1$ as below -
$4-x^2 \leq r \leq 5$ (distance from $y = 5 \ $ - $ \ $ i) to $y = 0$ which is $5$ and ii) to parabola $y = x^2 + 1$ which is $(5 - (x^2+1) = 4 - x^2$).
Also, $0 \leq x \leq 2$
So the integral is $ \displaystyle \int_0^{2\pi} \int_0^2 \int_{4 - x^2}^5 r \ dr \ dx \ d\theta = \frac{494 \pi}{15}$
Now coming to the integral that you have come up with is based on
$r = 5 - y, 0 \leq x \leq \sqrt{y-1}$
$V = \displaystyle \iiint r \ dx \ d\theta \ dr = \displaystyle \int_1^5 \int_0^{2\pi} \int_0^{\sqrt{y-1}} (5-y) \ dx \ d\theta \ dy$
$V = \displaystyle 2\pi \int_1^5 (5-y) \sqrt{y-1} \ dy = \frac{256 \pi}{15}$
Subtract it from the cylinder volume which is $\displaystyle \int_0^{2\pi} \int_0^5 \int_0^2 y \ dx \ dy \ d\theta = 50 \pi$
So volume of region $1 \displaystyle = 50 \pi - \frac{256 \pi}{15} = \frac{494 \pi}{15}$