I have a following integral: $$\int_0^1 dx\int_0^{\sqrt{1-x^2}}dy \int_\sqrt{x^2+y^2}^\sqrt{1-x^2-y^2}z^2dz$$

Which i have to solve by introducing polar coordinates, which is, by itself, relatively simple:

$$x=\rho\cos\theta\sin\phi \\ y=\rho\sin\theta\sin\phi \\ z=\rho\cos\phi$$

Besides this, i need to find Jacobian since i introduced a substitution, and since this is well known substitution Jacobian is $$J=\rho^2\sin\phi$$

Now, since i introduced polar coordinates, bounds of integral should be in polar form too, lower bound of the first, $dz$ integral, is simple $$\sqrt{x^2+y^2}= \rho\sin\phi$$, but i don't know what to do with this expression $$\sqrt{1-x^2}=\sqrt{1-\rho^2\cos^2\theta\sin^2\phi}$$

since, after introducing polar coordinates, this bound has all of the variables in itself, which makes it impossible to integrate over any of the variables i have, so i don't know how to solve this. Any help appreciated.

  • $\begingroup$ Hint: You should visualise the shape of the volume first. Then write down the bounds for spherical coordinates. Once you see what this volume is like, you can easily do the rest. $\endgroup$
    – user122049
    Dec 17, 2017 at 9:36
  • $\begingroup$ I suppose that then i should firstly visualize the area in a plane, which would be the upper semicircle with $x$ going from zero to one, and then, in order to construct the volume of the given set, i should move along the $z$ axis from one paraboloid to another since $\sqrt{x^2+y^2}$ and $\sqrt{1-x^2-y^2}$ represent two different paraboloids in three dimensional space, however, intuitively, intersection of two paraboloids calls for a cylindrical coordinates, but i have to use polar ones here and i still don't know how. $\endgroup$
    – cdummie
    Dec 17, 2017 at 11:01
  • $\begingroup$ Given $z=\sqrt{x^2+y^2}$, we can do $z^2 = x^2+y^2$. This is a cone. This is the way you can visualise the volume $\endgroup$
    – user122049
    Dec 18, 2017 at 10:24

1 Answer 1


Actually, this is a volume integral of the form $\iiint_V f(x,y,z) dx dy dz$ ; where V is the volume(can be define by the limits of the given triple integration).

Since you want to solve this by using polar co-ordinate system ,so you need to know the limits of $\rho$ ,$\theta$ and $\phi$. For this you have to understand about $\rho$=C(sphere of radius C and centre at origin) , $\phi$=C(pair of cone with semi-vertical angle C and $z-axis$ as its axis) and $\theta$=C(pair of straight lines passing through origin) ,C is any constant.

Here, $V$ is the volume in the $+ve$ part of $ z-axis$(only in the first octant) , enclosed between $\rho=1$ and $\phi$=$\pi/4$.

Therefore, the limits are-

$\rho$: $0\to1$

$\phi$: $0\to\pi/4$

$\theta$:$ 0\to \pi/2$

so, the integral can be given as follows,

$=$$\int \rho^4 d\rho$ $\int\sin\phi \cos^2\phi$ $d\phi$ $\int d\theta$.

After applying limits , you'll get the answer.

  • $\begingroup$ The amount of $\theta$ must be $0 \to \pi/2$. $\endgroup$
    – M.R. Yegan
    Dec 17, 2017 at 12:44
  • $\begingroup$ @M.R. Yegan you are right.. because the limits of $x$ and $y$ are taken in the first octant.., missed it.. $\endgroup$
    – user455480
    Dec 17, 2017 at 13:47
  • 1
    $\begingroup$ @cdummie , Either accept this answer or let me know your confusion.. $\endgroup$
    – user455480
    Dec 17, 2017 at 17:13
  • $\begingroup$ your answer to the problem is complete. $\endgroup$
    – M.R. Yegan
    Dec 17, 2017 at 19:06
  • $\begingroup$ @Ok i understand that $\rho=C$ is a sphere, but i have no idea what cone with semi-vertical angle is, and what is $+ve$ part of $z$-axis, what is $+ve$ anyway? $\endgroup$
    – cdummie
    Dec 18, 2017 at 18:20

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