# Calculating volume of spheroid using calculus 1

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

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?

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

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

• I made a mistake copying it over to the forum. I corrected above and I'm still stumped. – Sean Mahoney Jun 28 '18 at 1:47

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

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

• Thanks for the feedback. On paper I used the right form of 1 - x^2 / 4, I just copied it here wrong. Why is the volume smaller if you rotate around the x axis? The volume should be the same no matter which way you rotate it right? And yeah, the solution I posted does it the way you do it. I initially solved it by rotating it around x though, so was trying to do it that way. – Sean Mahoney Jun 28 '18 at 1:44
• The region is an ellipse with minor axis 1 and major axis 2. When you rotate you get an ellipsoid. But if you rotate around the y axis the 3rd axis of the ellipsoid will be 2. And if you rotate around the x axis the 3rd axis will be 1. The linked solution has $V = \pi \int_{-1}^1 x^2 \ dy = \pi \int_{-1}^1 4-4y^2 \ dy$ which gives the same result but looks rather different. – Doug M Jun 28 '18 at 2:15