Revolving the area between two functions around the y-axis

The question comes from the 1991 AP Calculus test.

Let $R$ be the region between the graphs of $y = 1 + sin(\pi x)$ and $y = x^2$ from $x = 0$ to $x = 1$.

c) Set up, but do not integrate an integral expression in terms of a single variable for the volume of the solid generated when R is revolved about the y-axis

$$V = 2\pi\int_{0}^{1}x(1 + sin(\pi x) - x^2)dx$$

or

$$V = \pi\int_{0}^{1}y dy + \pi\int_{1}^{2}(1 - \frac{1} \pi arcsin(y-1))^2 - (\frac{1} \pi arcsin(y-1))^2dy$$

I am confused on how they reached these two answers. I was able to do the question before this which involved rotating the same area but around the x-axis. In school, we learned to write the equations as $x = ...$ when dealing with rotations around the y-axis but I'm not sure how that applies here.

• The first formula uses the method that is probably called something like cylindrical shells. The second is more messy. We take lices perpendicular to the $y$-axis, and calculate the area of cross-section, and integrate with respect to $y$. The details of this one take a while to explain. Mar 26 '15 at 22:24
• The reason why the second method has two integrals is because you have a break when you switch from taking area under one function to area under the other. That uses the disk method, which is often easier over the x-axis, but tends to have more of these "switching function" points when taken over the y-axis. Mar 26 '15 at 23:01

1) Using the shell method, $\displaystyle V=\int_0^1 2\pi r(x)h(x)dx=\int_0^1 2\pi x(1+\sin(\pi x)-x^2)\;dx$

$\hspace{.2 in}$since $r(x)=x-0=x$ and $h(x)=y_1-y_2=1+\sin(\pi x)-x^2$.

2) Using the disc method, we can divide the region into 2 parts using the line segment from (0,1) to (1,1).

Then $V=\displaystyle\int_0^1 \pi((R_1(y))^2-(r_1(y))^2)\;dy+\int_1^2 \pi((R_2(y))^2-(r_2(y))^2)\;dy$ where

$\hspace{.4 in}y=x^2\implies R_1(y)=x=\sqrt{y}\;$ and $\;r_1(y)=0$.

Solving $y=1+\sin(\pi x)$ for $x$ gives $\sin(\pi x)=y-1$, so

a) $0\le x\le\frac{1}{2}\implies 0\le \pi x\le\frac{\pi}{2}\implies \pi x=\sin^{-1}(y-1)\implies x=\frac{1}{\pi}\sin^{-1}(y-1)$

$\displaystyle\hspace{3.4 in}\implies r_2(y)=\frac{1}{\pi}\sin^{-1}(y-1)$

b) $\frac{1}{2}\le x\le1\implies\frac{\pi}{2}\le \pi x\le \pi\implies \pi x=\pi-\sin^{-1}(y-1)$

$\displaystyle\hspace{2.05 in}\implies x=1-\frac{1}{\pi}\sin^{-1}(y-1)\implies R_2(y)=1-\frac{1}{\pi}\sin^{-1}(y-1)$

The first integral is called the cylindrical method. It finds the volume by adding larger and larger thinly sliced cylinders on top of each other...like those stacking Russian dolls if they were to fill the whole space.

The idea is to integrate over the expanding radius of the cylinders. $V=\pi r^2 h$ The radius in this integral is x, because it is the distance from the axis of rotation, the y-axis. The height of the cylinder is the difference in the two functions represents the height of each cylinder.

So if you integrate $V=2\pi\int rh dr$ you get $2\pi * \frac12 r^2 h$ which is $V=\pi r^2 h$ Hence the cylindrical method.

You integrate from 0 to 1 to get all of the radii from the y-axis. Let me know if this description was helpful.

One way to proceed is to evaluate the volume integral $V=\int_V dV$. For a volume of a body of revolution, cylindrical coordinates facilitates. Thus, for the first integral of the problem

\begin{align} V &=\int_V dV\\ &=\int_0^{2\pi} \int_0^{1} \int_{\rho^2}^{1+\sin(\pi \rho)} dz\rho d\rho d\phi\\ &=2\pi \int_0^{1}\left(1+\sin(\pi \rho)-\rho^2\right) \rho d\rho \end{align}

For the second integral of the problem, the order of integration changes and we find

\begin{align} V &=\int_V dV\\ &=\int_0^{2\pi} \int_0^{1} \int_{0}^{\sqrt{z}} \rho d\rho dz d\phi+\int_0^{2\pi} \int_1^{2} \int_{\frac{1}{\pi}\arcsin (z-1)}^{\left(1-\frac{1}{\pi}\arcsin (z-1)\right)} \rho d\rho dz d\phi\\ &=\pi \int_0^{1} zdz+\pi \int_1^2 \left(\left(1-\frac{1}{\pi}\arcsin (z-1)\right)^2-\left(\frac{1}{\pi}\arcsin (z-1)\right)^2\right)dz \end{align}

where the strict definition of $\arcsin(x)$ was carefully applied. This is due to the fact that for $-1\le x\le1$, we have $-\frac{\pi}{2}\le\arcsin(x)\le \frac{\pi}{2}$. For the part of the body of revolution in the region $\frac{\pi}{2}\le \pi \rho \le \pi$, $\pi \rho = \pi - \arcsin(z-1)$.

Volume enclosed between cylindrical shells

$$V = \pi\int_{0}^{1}(x_1 ^2 - x_2^2) dy$$ Next we write x in terms of y ( because rotation is with respect to y-axis) for each shell.

$$V = \pi\int_{0}^{1} y\; dy - (\frac{1} \pi \sin^{-1}(y-1))^2dy$$ ONLY. But I cannot explain the presence of the middle term when

$$F(y)= \sin^{-1}( (y - 1)/ \pi) ^2$$ is single valued.