I have a problem which I've had a look on "Maths Stack Exchange" and other resources to help, but still am stuck, so any help would be most appreciated.
My Problem:
- Set up a triple integral for the volume of the sphere $S_{R}$, where $$S_{R} = {(𝑥,𝑦,𝑧)∈ℝ^3|𝑥^{2}+𝑦^{2}+𝑧^{2}=𝑅^{2}}$$, with $R>0$ is the radius of the sphere.
Help:
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I don't understand what type of coordinates I'm meant to be using for this, as I chose to use spherical coordinates, but others seem to use cylindrical coordinates.
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I know that the triple integral will look eventually like $$\int\int\int_{S_{R}}dV$$is this correct, where my limits will be (using spherical coordinates): $${\int^{r}}{\int^{\theta}}{\int^{\phi}}dr\,d\theta\,d\phi$$
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Also I don't understand how to compute my limits, as I'm only given the radius $R>0$, and haven't been given any range for the $x,y,z$ values. As I'm just meant to set up the triple integral, will this mean I'm not meant to be computing the limits?
Any help with this would be really great, and thank you in advance for any help and further understanding of my problem.
Best Answer
This is by far easiest in spherical coordinates, as long as you use the correct volume element of $dr\,\, r \sin \theta \, dr\,\, r\,d\phi = r^2 \sin \theta\,dr\,d\theta\,d\phi$ rather than simply writing $dr\,d\theta\,d\phi$.
$$ \int_{r=0}^R\int_{\theta = 0}^{\pi} \int_{\phi=0}^\pi r^2\sin\theta\,dr\,d\theta\,d\phi = \\ 2\pi\int_{r=0}^R\int_{\theta = 0}^{\pi} r^2\sin\theta\,dr\,d\theta = \\ 2\pi\int_{r=0}^R \left[ -r^2\cos\theta \right]_{\theta = 0}^\pi dr = -2\pi\int_{r=0}^R r^2 \left[ \cos\pi - \cos 0\right]\,dr \\ = 4\pi \int_{r=0}^R r^2dr = \frac43\pi R^3{}{} $$
It is a bit tougher in cylindrical coordinates, with volume element $r\,dr\,d\phi\,dz$, because the limits on $z$ are less simple: $$ \int_{r=0}^R\int_{\phi = 0}^{2\pi} \int_{z=-\sqrt{R^2-r^2}}^{\sqrt{R^2-r^2}} r\,dr\,d\phi\,dz = \\ 2\int_{r=0}^R\int_{\phi = 0}^{2\pi} r\sqrt{R^2-r^2} \,dr\,d\phi =\\ 4\pi\int_{r=0}^Rr\sqrt{R^2-r^2} \,dr\,d\phi =\\ 4\pi \left[-\frac13(R^2-r^2)^{3/2} \right]_0^R \\ =\frac43 \pi R^3 $$ It is toughest of all in Cartesian coordinates, but still can be done without truly scary integrals.