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Find the volume bounded by the cylinder $x^2 + y^2 = 1$, the planes $x=0, z=0, z=y$ and lies in the first octant. (where x, y, and z are all positive)

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What do you know about this type of problem? What have you tried? –  Sammy Black Apr 1 '13 at 7:02

3 Answers 3

Or by using the polar coordinates instead as below: $$\int_{r=0}^1\int_{\theta=0}^{\pi/2}\int_{z=0}^{y\sin(\theta)}rdzd\theta dr=1/3$$

enter image description here

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+1 Polar is good here, and the graphic really helps! ++++ –  amWhy Apr 1 '13 at 14:44
@amWhy: Thanks Amy, however, I was late here. :-) –  Babak S. Apr 1 '13 at 17:34
I wondered...I keep my eye out for you! ;-) $\quad$ Good to see you! –  amWhy Apr 1 '13 at 17:36

Since you are in the first octant, $x$ runs between $0$ and $1$. For each $x$, then $y$ runs from $0$ up to the circle, $y=\sqrt{1-x^2}$. Finally, for each $x$ and $y$, $z$ runs from 0 up to the plane $z=y$. The volume is given by:

$$\int_{x=0}^1\int_{y=0}^{\sqrt{1-x^2}}\int_{z=0}^y dzdydx= \int_{x=0}^1\int_{y=0}^{\sqrt{1-x^2}}y dydx=\frac{1}{3}$$

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Use a triple integral. We're simply trying to find the volume$-$there's no density involved or anything$-$so the important question is just to find the correct bounds.

Let's examine the $x$ variable first. $x$ can move from $0$ to $1$, since it's constrained below by $0$ and cannot be greater than $1$ (otherwise $x^2 + y^2 > 1$).

Next, look at $y$. $y$ is bounded below by $0$ (must lie in the first octet) and above by $\sqrt{1-x^2}$ (so that $x^2 + y^2 \le 1$).

Finally, look at $z$. $z$ is bounded below by $0$ and above by $y$ (because of the plane $z=y$)

\begin{align*} \int_0^1 \int_0^{\sqrt{1-x^2}} \int_0^y 1 dz dy dx &= \int_0^1 \int_0^{\sqrt{1-x^2}} ydydx \\ &= \int_0^1 \frac{1}{2} (1-x^2) dx \\ &= \frac{1}{2} - \frac{1}{6} \\ &= \frac{1}{3} \end{align*}

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