How do I find the volume of $\cos x$ about the $y$-axis?

Question

Evaluate the volume formed by rotating the region bounded by $y=0$ and $y=\cos x$ for $\frac{\pi}{2}\le x\le\frac{3\pi}{2}$ about the y-axis.

I tried making $x=\cos^{-1}y$ then solving for $$\int_{{\frac{\pi}{2}}}^{\frac{3\pi}{2}}\cos^{-1}y\ dy$$ but clearly that is undefined. How do I solve this?

Answer 1

HINT

\begin{align*} V = \int_{\pi/2}^{3\pi/2}2\pi x|\cos(x)|\mathrm{d}x = -2\pi\int_{\pi/2}^{3\pi/2}x\cos(x)\mathrm{d}x \end{align*}

This might be helpful in order to understand such formula.

Answer 2

We break down the area trapped for $y=\cos x$ in between $\frac{\pi}{2} < x < \frac{3\pi}{2}$ into strips of each length $dx$

Now, the differential volume of the strip on revolution is $dV = 2\pi x |\cos(x)|dx$

Hence, the total volume

$$V = \int_{\pi/2}^{3\pi/2}2\pi x |\cos x| dx$$

Answer 3

One way would be to use cylindrical shells: $-2\pi\int_{\pi/2}^{3\pi/2}x\cos x\rm dx$, where the minus sign is because $\cos$ is negative on $(π/2,3π/2)$.

The integral can be solved by integration by parts: $ x\sin x=\int \sin x\rm dx+\int x\cos x\rm dx$, so $\int x\cos x \rm dx=x\sin x+\cos x+C$.