# Finding the volume of an object with 3 parameters. given an equation of ellipsoid, it has 3 parameters: x^2+4y^2+4z^2 leq 4

Finding the volume of an object with 3 parameters
I know how to find the volume of a sphere/ball around x-axis using:
$V=\pi {\int }_{a}^{b}{f}^{2}\left(x\right)dx$
Lets say if:
${x}^{2}+{y}^{2}={r}^{2}$
We do: $y=\sqrt{{r}^{2}-{x}^{2}}$
So now: $V=\pi {\int }_{-r}^{r}\left(\sqrt{{r}^{2}-{x}^{2}}{\right)}^{2}dx$
But the problem starts here:
Im given an equation of ellipsoid, it has 3 parameters:
${x}^{2}+4{y}^{2}+4{z}^{2}\le 4$
What do i do with the z parameter? How do i build y now? how does it fit to the equation by integrals of V?
I would like an explanation and not just a solution - because its homework.
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ticotaku86
Step 1
In finding the volume of a ball ${x}^{2}+{y}^{2}+{z}^{2}\le {r}^{2}$, you revolve the curve $y=\sqrt{{r}^{2}-{x}^{2}}$ about the y-axis. Notice that $y=\sqrt{{r}^{2}-{x}^{2}}$ is the equation you get by setting $z=0$ in the equation for the sphere.
You can do the same thing with the ellipsoid: set $z=0$ to get the equation for the boundary of the rugby ball in the (x,y)-plane, and solve for y.
${x}^{2}+4{y}^{2}=4\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}y=\frac{\sqrt{4-{x}^{2}}}{2}$
Step 2
Revolving this curve about the y-axis gives the volume,
$\pi {\int }_{-2}^{2}{\left(\frac{\sqrt{4-{x}^{2}}}{2}\right)}^{2}\phantom{\rule{thinmathspace}{0ex}}\mathrm{d}x=\frac{\pi }{4}{\int }_{-2}^{2}4-{x}^{2}\phantom{\rule{thinmathspace}{0ex}}\mathrm{d}x=\frac{8\pi }{3}$

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