# Find integration limits and compute integral

I have this problem: Find integration limits and compute the following integral.

$$\iiint_D(1-z)\,dx\,dy\,dz \\ D = \{\;(x, y, z) \in R^3\;\ |\;\; x^2 + y^2 + z^2 \le a^2, z\gt0\;\}$$

I can compute this as an indefinite integral but finding integration limits beats me. As indefinite integral the result looks like this (hopefully without any careless mistakes):

$$\iiint(1-z)\,dx\, dy\, dz \\ = \iint(x(1-z) + C_x)\,dy\, dz\\ = \iint (x - xz + C_x)\,dy\, dz \\ = \int (xy - xyz + yC_x + C_y)\,dz\\ = xyz - \frac{xyz^2}{2} + yzC_x + zC_y + C_z$$

• change to spherical coordinates – i. m. soloveichik Aug 31 '12 at 15:37
• The region you are integrating over is a half ball. Are you familiar with spherical coordinates? Use the change of variables technique and the rest will be easy. – Ayman Hourieh Aug 31 '12 at 15:37
• @Ayaman I don't know about spherical coordinates. Well I just formed an intuition starting from the name. – Dr.Optix Aug 31 '12 at 16:05
• @Dr.Optix - The formulae are the end of the "Cartesian coordinates" section in this article: en.wikipedia.org/wiki/… – Ayman Hourieh Aug 31 '12 at 16:21

Change the limits as $$8*\int_0^a \int_0^{\sqrt{a^2 - x^2}} \int_0^{\sqrt{a^2 - x^2 - y^2}} (1 - z) \,dz \,dy \,dx$$ This integral should give you the volume of sphere. Note that it is easy to remember the limits. $$8*\int_0^a \int_0^{\sqrt{a^2 - x^2}} \int_0^{\sqrt{a^2 - x^2 - y^2}} 1 \cdot \,dz \,dy \,dx = {4 \over 3} \pi a^3$$

EDIT:: for $z > 0$ we have $$4*\int_0^a \int_0^{\sqrt{a^2 - x^2}} \int_0^{\sqrt{a^2 - x^2 - y^2}} (1 - z) \cdot \,dz \,dy \,dx$$

More exactly, the limits of integration are $$\int_{-a}^a \int_{-\sqrt{a^2 - x^2}}^{\sqrt{a^2 - x^2}} \int_0^{\sqrt{a^2 - x^2 - y^2}} (1 - z) \cdot \,dz \,dy \,dx$$

This can be easily be done by changing into spherical polar coordinates, $z = a \cos \theta$ and the limit of $\theta = 0 - {\pi \over 2}$.

• Thanx for the answer I will try to fully understand it – Dr.Optix Aug 31 '12 at 18:41
• you are welcome ... this is just like finding the area of the cirlce $$4 \int_0^a\int_0^{\sqrt{a^2 - x^2}} 1 \cdot dx \;dy$$ – Santosh Linkha Aug 31 '12 at 18:45
• except the fact that the circle varies with $z$ and $\sqrt{x^2+y^2}$ is your variable radius – Santosh Linkha Aug 31 '12 at 18:47

Use linearity: $$\int_D (1-z) \,\mathrm{d}V = \int_D 1 \,\mathrm{d}V - \int_D z \,\mathrm{d}V =\frac{2 \pi}{3} a^3 - \int_D z \,\mathrm{d}V$$ The latter integral is best evaluated in cylindrical coordinates $(x,y,z) = (r \sin(\phi), r \cos(\phi), z)$. The measure $dV$ factors into $dz$ times the measure on the disk $x^2+y^2 \leqslant a^2-z^2$, whose area is $\pi (a^2-z^2)$, thus $$\int_D z\, \mathrm{d}V = \int_0^a z \left( \pi (a^2-z^2) \right) \,\mathrm{d} z = \frac{\pi}{4} a^4$$

• Thanx for the answer I will try to fully understand it – Dr.Optix Aug 31 '12 at 18:42

Another approach, with cylindrical coordinates: $$x=r\cos\theta\,\,,\,\,y=r\sin\theta\,\,,\,\,z=z\Longrightarrow |J|=r\,\, (\,\text{Jacobian})$$ Since we want to integrate on the uper half sphere $\,x^2+y^2+z^2\leq a^2\,\,,\,\,z\geq 0$ , we get: $$0\leq r\leq a\,\,,\,\,0\leq\theta\leq 2\pi\,\,,\,\,0\leq z\leq \sqrt {a^2-x^2-y^2}=\sqrt {a^2-r^2}$$ so our integral becomes $$\int_0^adr\int_0^{2\pi}d\theta\int_0^\sqrt{a^2-r^2}r(1-z)dz=2\pi\int_0^ar\,dr\left.\left(z-\frac{1}{2}z^2\right)\right|_0^\sqrt{a^2-r^2}=$$ $$=-\pi\int_0^a(-2r\,dr)\sqrt{a^2-r^2}-\pi\int_0^a\left(a^2r-r^3\right)dr=$$ $$=\left.-\frac{2\pi}{3}\left(a^2-r^2\right)^{3/2}\right|_0^a-\left.\pi\left(\frac{a^2r^2}{2}-\frac{r^4}{4}\right)\right|_0^a=\frac{2\pi}{3}a^3-\frac{\pi}{4}a^4$$