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The question actually asks to solve for c where the integral is $\iint\limits_D c\sqrt{1-x^2-y^2}dA = 1$, where $D$ is the disk $x^2+y^2\le1$. Anyways, for the integral, I thought I should use polar coordinates so I substituted $x^2+y^2$ with $r^2$. When I do this, I would get $\iint\limits_D \sqrt{1-r^2}drd\theta$ where $D$ will now be represented as $0\le{r}\le{1}$ and $0\le{\theta}\le{2\pi}$. But I'm not sure...I'm pretty rusty with my calculus so please help me out.

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When you go to polar, $dy\,dx$ becomes $r\,dr\,d\theta$. –  Gerry Myerson Mar 7 '13 at 1:37

3 Answers 3

up vote 1 down vote accepted

Actually, $dA = r dr d\theta$, so you get

$$c \cdot 2 \pi \int_0^1 dr \, r \sqrt{1-r^2} = 1$$

Note that

$$\int_0^1 dr \, r \sqrt{1-r^2} = \frac{1}{2} \int_0^1 du \sqrt{1-u} = \frac{1}{3}$$

$$\therefore \: c = \frac{3}{2 \pi}$$

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I appreciate this, thank you!!! –  Kiyoshi Mar 7 '13 at 2:06

Almost right. When you make the change of variable, $dx\,dy$ should be replaced by $r\,dr\,d\theta$.

The $r\,dr\,d\theta$ can be thought of as the Jacobian of the transformation. Informally, $dx\,dy$ is the area of a little $dx$ by $dy$ rectangle with one corner at $(x,y)$, and $r\,dr\,d\theta$ approximates, in polar terms, the area of that rectangle.

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Thank you!!! I have to go review my calculus. Your answer was very helpful :) –  Kiyoshi Mar 7 '13 at 2:06
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It is often covered in first-year calculus, in a section on area, arclength in polar coordinates. Then the general idea, with transformations that go beyond the rectangular to polar, is done in the several variables course. I mentioned the Jacobian so you would have a term to search for, though "area in polar coordinates" will work for your problem. –  André Nicolas Mar 7 '13 at 2:11

This problem has a nice geometrical solution. Notice that the function under the integral is a hemisphere with radius $1$ whose intersection with the $XOY$ plain is $D$: $$z=\sqrt{1-x^2-y^2} \Longleftrightarrow x^2+y^2+z^2=1, z\ge0$$ Therefore the integral without the constant represents the volume of the hemisphere $$C\frac{2\pi}{3}=1\Longrightarrow C=\frac{3}{2\pi}.$$

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