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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 $x,y \ge 0$.

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    Is this homework? Could you explain what the problem is? Where did you get stuck?2011-10-25
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    Can you rephrase the question, so it sounds like a question and not an order?2011-10-25
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    @J.M.: I’ll not change it back, but I much prefer $dV$ to $\mathrm dV$.2011-10-25
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    @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.2011-10-25
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    @J.M.: Purely a matter of taste: I think that it looks much better, and I don’t consider the $d$ an operator.2011-10-25
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    Surely this would be easier in cylindrical coordinates?2011-10-25

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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

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

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    how can we find limit z(1-x^2)^1/2 w.r to x2011-10-25
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    Do a trig substitution with $x=\sin\theta$, $dx=\cos\theta\,d\theta$. Then $\sqrt{1-x^2}=\cos\theta$.2011-10-25
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Isn't the answer simply $\pi/4$ ? It is a quarter cylinder of unit height and unit radius, right?

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    No: the integrand is $z$, not $1$. However, it *is* simply $\frac{\pi}{4}\int_0^1z\;dz$.2011-10-25
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    Absolutely correct. I forgot the integrand...2011-10-25
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    can you explain how can we find integral z(1-x^2)1/2 w.r.to x2011-10-25
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    Now that you have seen the mistake, perhaps you should correct the answer incorporating @Brian's comment.2011-10-26