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How is trigonometric substitution done with a triple integral? For instance,

$$ 8 \int_0^r \int_0^{\sqrt{r^2-x^2}} \int_0^{\sqrt{r^2-x^2-y^2}} (1) dz dy dx $$

Here the limits have been chosen to slice an 8th of a sphere through the origin of radius r, and to multiply this volume by 8. Without converting coordinates, how might a trig substitution be done to solve this?

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2 Answers

"The hard way" to go at this multiple integral looks as follows:

The innermost integral has the value $\sqrt{r^2-x^2-y^2}$. The next integral we are faced with is $$I(x):=\int_0^{\sqrt{r^2-x^2}}\sqrt{r^2-x^2-y^2}\ dy\ .$$ During the integration $x$ is constant. "Trigonometric substitution" means here that we somehow should use $1-\sin^2 t \equiv\cos^2 t$ to get rid of the square root. Therefore we substitute $$y:=\sqrt{r^2-x^2} \sin t\ ,\quad dy= \sqrt{r^2-x^2}\cos t\ dt \qquad(0\leq t\leq {\pi\over2})\ ,$$ and $I(x)$ becomes $$I(x)=(r^2-x^2)\ \int_0^{\pi/2} \cos^2 t\ dt\ .$$ Now use the rule "$\cos^2(\omega t)$ or $\sin^2(\omega t)$ integrated over an integer number of quarter periods gives half of the length of the integration interval" and obtain $I(x)={\pi\over4}(r^2-x^2)$.

It remains to compute the outermost integral: $${\rm vol}(B_r)=8\int_0^r I(x)\ dx=2\pi\ \int_0^r(r^2-x^2)\ dx=2\pi (r^2x-{x^3\over 3})\Bigr|_0^r ={4\pi\over3}\ r^3\ .$$

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The volume of the sphere $B(0,r)=\{(x,y,z): x^2+y^2+z^2 \leq r^2\}$ is usually calculated as follows: Make the change of variable $x=r\cos \theta \sin \phi;\ y=r\sin \theta \sin \phi;\ z=r \cos \phi$, with the Jacobian equal to $r^2 \sin\phi$.

$V=\int_{B(0,r)}1 dx=\int_0^r \int_0^{2\pi} \int_0^\pi r^2 \sin \phi d \phi d \theta d r=\frac{4\pi r^3}{3}$.

For further reference on spherical coordinates, take a look at this article

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I understand the switch to spherical coordinates, the question is geared toward multi-variate trig subs. –  Jon Jun 1 '11 at 8:58
    
I don't understand your motivation. Why do you want to calculate the integral in a harder way? –  Beni Bogosel Jun 1 '11 at 9:15
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