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 April 7th, 2017, 05:06 PM #1 Senior Member   Joined: Jan 2017 From: Toronto Posts: 209 Thanks: 3 Double/Triple Integral for volume between surfaces Find the volume below z = 1 - y above the region -1 <= x <= 1, 0 <= y <= 1 - x^2. Answer: 4/5 Any tips would be apprecipated!
 April 7th, 2017, 05:13 PM #2 Senior Member     Joined: Sep 2015 From: USA Posts: 2,590 Thanks: 1434 I presume you also mean $z\geq 0$ \begin{align*} \displaystyle \int_{-1}^1 \int_0^{1-x^2}\int_0^{1-y} ~dz~dy~dx \end{align*} there shouldn't be anything tricky about this integral
 April 8th, 2017, 05:23 AM #3 Senior Member   Joined: Jan 2017 From: Toronto Posts: 209 Thanks: 3 Thanks! May I ask another similar question Find the volume bounded by z = x^2 + y^2 and z = 4 and z = 0 Answer: 8pi I tried to use the same trick but couldn't get the correct answer. \begin{align*} \displaystyle \int_{-2}^2 \int_{-2}^2\int_0^{x^2+y^2} ~dz~dy~dx \end{align*} Last edited by zollen; April 8th, 2017 at 05:26 AM.
 April 8th, 2017, 07:08 AM #4 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,264 Thanks: 902 Your bounds on x and y are incorrect. $-2\le x\le 2$, $-2\le y\le 2$, in the xy-plane, is a square. z= 4 corresponds to the circle $x^2+ y^2= 4$. You can cover that by taking $-2\le x\le 2$ and, for each x, $-\sqrt{4- x^2}\le y\le \sqrt{4- x^2}$. A simpler thing to do might be to use polar coordinates in the xy-plane. Then the bounds would be $0\le r\le 2$, $0\le\theta\le 2\pi$. (The "differential of area" in polar coordinates is $rdrd\theta$.) Last edited by Country Boy; April 8th, 2017 at 07:13 AM.
April 8th, 2017, 10:04 AM   #5
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Dang! You are correct!

Quote:
 Originally Posted by Country Boy Your bounds on x and y are incorrect. $-2\le x\le 2$, $-2\le y\le 2$, in the xy-plane, is a square. z= 4 corresponds to the circle $x^2+ y^2= 4$. You can cover that by taking $-2\le x\le 2$ and, for each x, $-\sqrt{4- x^2}\le y\le \sqrt{4- x^2}$. A simpler thing to do might be to use polar coordinates in the xy-plane. Then the bounds would be $0\le r\le 2$, $0\le\theta\le 2\pi$. (The "differential of area" in polar coordinates is $rdrd\theta$.)

April 8th, 2017, 12:39 PM   #6
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I have been considering your polar coordinates solution. Correct me if I am wrong. But your integral does not seem to cover the z-axis where 0 <= z <= 4

$rdrd\theta$.

Quote:
 Originally Posted by Country Boy A simpler thing to do might be to use polar coordinates in the xy-plane. Then the bounds would be $0\le r\le 2$, $0\le\theta\le 2\pi$. (The "differential of area" in polar coordinates is $rdrd\theta$.)

 April 9th, 2017, 03:40 AM #7 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,264 Thanks: 902 r= 0 is the z-axis. The volume is given by $\int_0^{2\pi}\int_0^2 r^2 (rdrd\theta)= \int_0^{2\pi}\int_0^2 r^3 dr d\theta$. Using x, y, z coordinates you would have $\int_{-2}^2\int_{-\sqrt{4- x^2}}^{\sqrt{4- x^2}} x^2+ y^2 dydx$. You should get the same answer for both of those. Thanks from zollen
April 9th, 2017, 08:39 AM   #8
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May I ask how did you get the r^2?

$\int_0^{2\pi}\int_0^2 r^2 (rdrd\theta)$

Quote:
 Originally Posted by Country Boy r= 0 is the z-axis. The volume is given by $\int_0^{2\pi}\int_0^2 r^2 (rdrd\theta)= \int_0^{2\pi}\int_0^2 r^3 dr d\theta$. Using x, y, z coordinates you would have $\int_{-2}^2\int_{-\sqrt{4- x^2}}^{\sqrt{4- x^2}} x^2+ y^2 dydx$. You should get the same answer for both of those.

 April 9th, 2017, 09:51 AM #9 Senior Member   Joined: Jan 2017 From: Toronto Posts: 209 Thanks: 3 after much contemplation, I realized that r^2=x^2+y^2

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