Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question

"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: $${\rm vol}(B_r)=8\int_0^r I(x)\ dx=2\pi\ \int_0^r(r^2-x^2)\ dx=2\pi (r^2x-{x^3\over 3})\Bigr|_0^r ={4\pi\over3}\ r^3\ .$$

share|cite|improve this answer

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

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.