Calculating volume of spheroid using calculus 1

Question

I'm refreshing my calculus knowledge with MIT Open courseware, and cannot figure out how to solve this problem using a slightly different method:

https://ocw.mit.edu/courses/mathematics/18-01sc-single-variable-calculus-fall-2010/unit-3-the-definite-integral-and-its-applications/part-b-second-fundamental-theorem-areas-volumes/session-58-volume-of-a-sphere-revolving-about-x-axis/MIT18_01SCF10_ex58sol.pdf

They solve it by slicing it up in the y direction, but I tried it in the x direction and have a mistake somewhere. I ended up with:

$$y^2 = \frac{4-x^2}{4} = 1- \frac{x^2}{4}$$ $$dV = \int_2^{-2}\pi y^2 dx$$ $$dV = \int_2^{-2}\pi 1 - \frac{x^2}{4} dx$$ $$\pi\int_2^{-2} x - \frac{x^3}{12} dx = \frac{8}{3} \pi$$

Have I made a simple arithmetic error or is the integral formula not setup correctly?

Answer 1

I think the Integrand musst be $$\frac{1}{4}-\frac{x^2}{4}$$ and so we get

$$[\frac{x}{4}-\frac{x^3}{12}]_{-2}^2$$

Answer 2

Note: $\frac {4- x^2}{4} = 1 - \frac{x^2}{4}$, not $\frac{1 -x^2}{4}.$ It looks like you have corrected for this in the last line.

But your bigger issue one, you have rotated around the $x$ axis, and so the volume is smaller in this direction of revolution.

You can slice in the $y$ direction and revolve around the $y$ axis. But you will need to use cylindrical shells and not disks.

i.e. $y = \pm \sqrt {1 - \frac {x^2}{4}} = \pm \frac {\sqrt {4 - x^2}}{2}$

$V = 2\pi \int_0^2 xy\ dx = 2\pi \int_0^2 x\sqrt {4-x^2}\ dx$