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April 7th, 2017, 05:06 PM   #1
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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!
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April 7th, 2017, 05:13 PM   #2
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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
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April 8th, 2017, 05:23 AM   #3
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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.
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April 8th, 2017, 07:08 AM   #4
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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.
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April 8th, 2017, 10:04 AM   #5
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Dang! You are correct!

Quote:
Originally Posted by Country Boy View Post
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 .)
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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 View Post
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 .)
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April 9th, 2017, 03:40 AM   #7
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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
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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 View Post
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.
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April 9th, 2017, 09:51 AM   #9
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after much contemplation, I realized that r^2=x^2+y^2
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