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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 $z\geq 0$ $\begin{align*} \displaystyle \int_{1}^1 \int_0^{1x^2}\int_0^{1y} ~dz~dy~dx \end{align*}$ 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 xyplane, 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 xyplane. 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  
Senior Member Joined: Jan 2017 From: Toronto Posts: 209 Thanks: 3 
Dang! You are correct! Quote:
 
April 8th, 2017, 12:39 PM  #6 
Senior Member Joined: Jan 2017 From: Toronto Posts: 209 Thanks: 3 
I have been considering your polar coordinates solution. Correct me if I am wrong. But your integral does not seem to cover the zaxis where 0 <= z <= 4 . 
April 9th, 2017, 03:40 AM  #7 
Math Team Joined: Jan 2015 From: Alabama Posts: 3,264 Thanks: 902 
r= 0 is the zaxis. 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, 08:39 AM  #8 
Senior Member Joined: Jan 2017 From: Toronto Posts: 209 Thanks: 3  
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 
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