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 April 10th, 2017, 04:19 PM #1 Senior Member   Joined: Jan 2017 From: Toronto Posts: 175 Thanks: 2 Double Integration about general equation Find the volume bounded by z = x^2 + y^2 and z = y. Answer: pi/32 Any tips would be much appreciated.
 April 11th, 2017, 03:32 AM #2 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,165 Thanks: 867 There is no area bounded by these. The first is a cylinder, the second is a plane cutting the cylinder. You need another plane to bound a finite region. Was there a condition that z be positive? That is, that the xy-plane is also a boundary?
April 11th, 2017, 09:38 AM   #3
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Quote:
 Originally Posted by Country Boy That is, that the xy-plane is also a boundary?
that's what I would assume

 April 12th, 2017, 05:25 AM #4 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,165 Thanks: 867 I wouldn't. Hopefully, Zollen will get back to us to tell us what the problem really says!
 April 12th, 2017, 02:41 PM #5 Senior Member   Joined: Jan 2017 From: Toronto Posts: 175 Thanks: 2 Yes. z must be positive.
 April 13th, 2017, 02:36 AM #6 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,165 Thanks: 867 Apparently I did not look closely enough at the original problem! $z= x^2+ y^2$ is a paraboloid, not a cylinder, and together with z= y does define a finite region. Taking z= y, the intersection of the plane and the paraboloid satisfies $y= x^2+ y^2$ or $x^2+ y^2- y= 0$. Complete the square in y by adding 1/4 to both sides: $x^2+ y^2- y+ 1/4= x^2+ (y- 1/2)^2= 1/4$. That projects to the circle in the xy plane with center at (0, 1/2) and radius 1/2. We can cover that by taking x from -1/2 to 1/2 and, for each x, taking y from $1/2- \sqrt{1/4- x^2}$ to $1/2+ \sqrt{1/4- x^2}$. The volume is given by $\int_{-1/2}^{1/2}\int_{1/2- \sqrt{1/4- x^2}}^{1/2+ \sqrt{1/4- x^2}} (y- x^2- y^2) dydx$ Thanks from topsquark and zollen Last edited by Country Boy; April 13th, 2017 at 02:39 AM.

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