# Evaluate the triple integral $\iiint\limits_D z \ dV$ over this domain $D$

Evaluate the integral $$\iiint \limits_D z \ dV ,$$ where $D$ is the region bounded by the planes $y = 0$, $x = 0$, $z = 0$, $z = 1$, and the cylinder $x^2+y^2=1$ with $x,y \ge 0$.

-
Is this homework? Could you explain what the problem is? Where did you get stuck? –  abatkai Oct 25 '11 at 21:41
Can you rephrase the question, so it sounds like a question and not an order? –  Gerry Myerson Oct 25 '11 at 21:43
@J.M.: I’ll not change it back, but I much prefer $dV$ to $\mathrm dV$. –  Brian M. Scott Oct 25 '11 at 21:47
@Brian: Knuth's argument for a Roman-type d is still ringing in my brain. If you've got a nice argument for not doing so for multiple integrals, I'll gladly undo. –  Ｊ. Ｍ. Oct 25 '11 at 22:03
Surely this would be easier in cylindrical coordinates? –  Mike Spivey Oct 25 '11 at 22:34

Usually the toughest part of these problems is finding the limits of integration. If we concentrate on just the $xy$-plane for a moment we can find the limits of integration for $x$ and $y$. The region in the $xy$-plane over which you are integrating is the region bounded by the circle $x^2+y^2=1$ in the first quadrant. Along the $x$-axis this region runs from $x=0$ to $x=1$. If we pick a particular $x$, then $y$ will run from $y=0$ to $y=\sqrt{1-x^2}$. So, now we've got bounds on $x$ and $y$. As for $z$, that certainly runs from $z=0$ to $z=1$. This gives all the bounds and the integral is

$$\int_0^1\int_0^1\int_0^{\sqrt{1-x^2}}z\,dy\,dx\,dz.$$

-
how can we find limit z(1-x^2)^1/2 w.r to x –  user18248 Oct 25 '11 at 22:27
Do a trig substitution with $x=\sin\theta$, $dx=\cos\theta\,d\theta$. Then $\sqrt{1-x^2}=\cos\theta$. –  Joe Johnson 126 Oct 25 '11 at 22:39

Isn't the answer simply $\pi/4$ ? It is a quarter cylinder of unit height and unit radius, right?

-
No: the integrand is $z$, not $1$. However, it is simply $\frac{\pi}{4}\int_0^1z\;dz$. –  Brian M. Scott Oct 25 '11 at 21:49
+1 for the answer with the comment adjoined –  Joe Johnson 126 Oct 25 '11 at 22:00
Absolutely correct. I forgot the integrand... –  Gandhi Viswanathan Oct 25 '11 at 22:02
can you explain how can we find integral z(1-x^2)1/2 w.r.to x –  user18248 Oct 25 '11 at 22:25
Now that you have seen the mistake, perhaps you should correct the answer incorporating @Brian's comment. –  Srivatsan Oct 26 '11 at 2:14