How to calculate the volume of $\cos(x)$ around the x-axis and the y-axis separately?

Question

I have difficulties finding the right formula how to calculate the volume rotating $\cos(x)$ around the x-axis and y-axis separately can you please give me a hint how to do it?

the interval is $x=[-\frac{\pi}{2},\frac{\pi}{2}]$

Thank you

Answer 1

HINT : Draw the figure, and you'll see the reason of the following formula :

1) $x$-axis case : $$\pi\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}y^2dx$$ where $y=\cos x.$

2) $y$-axis case : $$\pi\int_{0}^{1}x^2dy.$$ Do you know how to calculate this?

Answer 2

First draw the function. We know that Volume is given by $\int_a^b \pi f(x)^2 dx$. Therefore $\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\pi \cos^2(x) dx$ and that's an easy integral.

While integrating along y axis your function will change to $\cos^{-1}x $ and hence integral will become $\int_0^1 \pi (\cos^{-1}(y))^2 dy$.

Answer 3

1) Around the x-axis, you get $\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \pi (\cos x)^2dx=2\int_{0}^{\frac{\pi}{2}} \pi (\cos x)^2dx$,

$\;\;\;$using the Disc method (and symmetry).

2) Around the y-axis, you get $\int_{0}^{\frac{\pi}{2}}2\pi x\cos x\; dx$, using the Shell method.

$\;\;\;$(Notice that the right half of the region generates the whole solid in this case.)