How do I use polar coordinates to find the volume of a sphere of radius $r$ $r$ ?

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Answer 1 · 2015-09-10T23:00:31

The volume is $V = \frac{4}{3} π r^{3}$ $V=4/3pir^3$

The equation of a sphere is $x^{2} + y^{2} + z^{2} = r^{2}$ $x^2+y^2+z^2=r^2$

From the equation we get

$z = \pm \sqrt{r^{2} - (x^{2} + y^{2})}$ $z=+-sqrt(r^2-(x^2+y^2)$

The volume of the sphere is given by

$V = 2 \int \int_{x^{2} + y^{2} \leq r}^{2} \sqrt{r^{2} - x^{2} - y^{2}} d A$ $V=2intint_(x^2+y^2<=r)sqrt(r^2-x^2-y^2)dA$

Using polar coordinates $x = r cos a, y = r sin a$ $x=rcosa, y=rsina$ and substituing

to the integral above

$V = 2 \int_{0}^{2 \cdot π} \int_{0}^{r} \sqrt{r^{2} - a^{2}} r d r d a$ $V=2int_0^(2*pi)int_0^rsqrt(r^2-a^2)rdrda$

Which is calculated easily giving $V = \frac{4}{3} π r^{3}$ $V=4/3pir^3$

How do I use polar coordinates to find the volume of a sphere of radius rr?