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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. , , in the xy-plane, is a square. z= 4 corresponds to the circle . You can cover that by taking and, for each x, . A simpler thing to do might be to use polar coordinates in the xy-plane. Then the bounds would be , . (The "differential of area" in polar coordinates is .) 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. , , in the xy-plane, is a square. z= 4 corresponds to the circle . You can cover that by taking and, for each x, . A simpler thing to do might be to use polar coordinates in the xy-plane. Then the bounds would be , . (The "differential of area" in polar coordinates is .) 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

.

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 , . (The "differential of area" in polar coordinates is .) 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 . Using x, y, z coordinates you would have . 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?

Quote:
 Originally Posted by Country Boy r= 0 is the z-axis. The volume is given by . Using x, y, z coordinates you would have . 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 Tags double or triple, integral, surfaces, volume Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post Jwilks93 Calculus 5 March 10th, 2016 01:57 AM Aftermath Calculus 1 May 11th, 2015 01:59 AM triplekite Calculus 1 December 5th, 2012 08:16 AM PedroMinsk Calculus 3 December 9th, 2010 03:59 AM jim626 Calculus 0 May 4th, 2008 05:19 PM

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