Question #2602

Use polar coordinate to find the volume of solid inside the cylinder and inside the ellipsoid.

Expert's answer

1) Suppose that one of the base D of the cylinder is a disk of radius R in the plane xy and centered at the origin.

Let also its height is equal to H. Then its volume can be calculated as integral

<img src="/cgi-bin/mimetex.cgi?%5Cint_D%20Hr%20dr%20d%5Cphi%20=%20H%20%5Cint_0%5ER%20r%20dr%20%5Cint_0%5E%7B2%5Cpi%7Dd%5Cphi%20=%20H%20%5Cfrac%7Br%5E2%7D%7B2%7D%202%20%5Cpi%20=%20%5Cpi%20R%5E2%20H" title="\int_D Hr dr d\phi = H \int_0^R r dr \int_0^{2\pi}d\phi = H \frac{r^2}{2} 2 \pi = \pi R^2 H">

2) In general, ellipsoid is given by equation

<img src="http://latex.codecogs.com/gif.latex?%5Cfrac%7Bx%5E2%7D%7Ba%5E2%7D+%20%5Cfrac%7By%5E2%7D%7Bb%5E2%7D%20+%20%5Cfrac%7Bz%5E2%7D%7Bc%5E2%7D%20=%201" title="\frac{x^2}{a^2}+ \frac{y^2}{b^2} + \frac{z^2}{c^2} = 1">

The simples way to find its volume via polar coordinates is first assume that a=b=c=1, so our ellipsoid is a ball of radius 1 centered at the origin.

Moreover, we can also calculate only the volume of the art of this ball over the plane xy.

<img src="http://latex.codecogs.com/gif.latex?V%20=%202%20%5Cint_D%20%5Csqrt%7B1-x%5E2-y%5E2%7D%20dxdy" title="V = 2 \int_D \sqrt{1-x^2-y^2} dxdy">

where D is the unit disk in the plane xy centered at origin.

Then in polar coordinates we will have:

<img src="http://latex.codecogs.com/gif.latex?V%20=%202%20%5Cint_D%20%5Csqrt%7B1-r%5E2%7Drdr%20d%5Cphi%20=%202%20%5Cint_0%5E1%20%5Csqrt%7B1-r%5E2%7D%20rdr%20%5Cint_0%5E%7B2%5Cpi%7Dd%20%5Cphi%20=%20%5Cfrac%7B4%7D%7B3%7D%20%5Cpi" title="V = 2 \int_D \sqrt{1-r^2}rdr d\phi = 2 \int_0^1 \sqrt{1-r^2} rdr \int_0^{2\pi}d \phi = \frac{4}{3} \pi">

Now ellipsoid can be obtained from the ball by extension along x-axis in a-times, along y-axis in b times, and along z-axis in c-times.

This multiplies that volumes by abc.

Hence the volume of the ellipsoid is

<img src="http://latex.codecogs.com/gif.latex?%5Cfrac%7B4%7D%7B3%7D%5Cpi%20abc" title="\frac{4}{3}\pi abc">

Let also its height is equal to H. Then its volume can be calculated as integral

<img src="/cgi-bin/mimetex.cgi?%5Cint_D%20Hr%20dr%20d%5Cphi%20=%20H%20%5Cint_0%5ER%20r%20dr%20%5Cint_0%5E%7B2%5Cpi%7Dd%5Cphi%20=%20H%20%5Cfrac%7Br%5E2%7D%7B2%7D%202%20%5Cpi%20=%20%5Cpi%20R%5E2%20H" title="\int_D Hr dr d\phi = H \int_0^R r dr \int_0^{2\pi}d\phi = H \frac{r^2}{2} 2 \pi = \pi R^2 H">

2) In general, ellipsoid is given by equation

<img src="http://latex.codecogs.com/gif.latex?%5Cfrac%7Bx%5E2%7D%7Ba%5E2%7D+%20%5Cfrac%7By%5E2%7D%7Bb%5E2%7D%20+%20%5Cfrac%7Bz%5E2%7D%7Bc%5E2%7D%20=%201" title="\frac{x^2}{a^2}+ \frac{y^2}{b^2} + \frac{z^2}{c^2} = 1">

The simples way to find its volume via polar coordinates is first assume that a=b=c=1, so our ellipsoid is a ball of radius 1 centered at the origin.

Moreover, we can also calculate only the volume of the art of this ball over the plane xy.

<img src="http://latex.codecogs.com/gif.latex?V%20=%202%20%5Cint_D%20%5Csqrt%7B1-x%5E2-y%5E2%7D%20dxdy" title="V = 2 \int_D \sqrt{1-x^2-y^2} dxdy">

where D is the unit disk in the plane xy centered at origin.

Then in polar coordinates we will have:

<img src="http://latex.codecogs.com/gif.latex?V%20=%202%20%5Cint_D%20%5Csqrt%7B1-r%5E2%7Drdr%20d%5Cphi%20=%202%20%5Cint_0%5E1%20%5Csqrt%7B1-r%5E2%7D%20rdr%20%5Cint_0%5E%7B2%5Cpi%7Dd%20%5Cphi%20=%20%5Cfrac%7B4%7D%7B3%7D%20%5Cpi" title="V = 2 \int_D \sqrt{1-r^2}rdr d\phi = 2 \int_0^1 \sqrt{1-r^2} rdr \int_0^{2\pi}d \phi = \frac{4}{3} \pi">

Now ellipsoid can be obtained from the ball by extension along x-axis in a-times, along y-axis in b times, and along z-axis in c-times.

This multiplies that volumes by abc.

Hence the volume of the ellipsoid is

<img src="http://latex.codecogs.com/gif.latex?%5Cfrac%7B4%7D%7B3%7D%5Cpi%20abc" title="\frac{4}{3}\pi abc">

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