Calc 3 Double Integrals Problem

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hxnoka

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(I am mainly stuck on how to write the integral given this information, but would appreciate a fully solved solution too.)
Thank you.
 
Subhotosh Khan said:
What are the methods that you have been taught?
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For the chapter that I'm on, I have been taught solving double & triple integrals with x-simple and y-simple regions, changing the order of integration, and using u-substitutions. I'm not sure if that exactly applies to this problem.
 
If f(x)=z=1−x2+y2\displaystyle f(x)= z= 1- \sqrt{x^2+y^2} then z−1=−x2+y2\displaystyle z- 1=-\sqrt{x^2+ y^2} and (z−1)2=x2+y2\displaystyle (z- 1)^2= x^2+ y^2. That is a cone with center at (0, 0, 1). You are asked to find the volume of the nappe below (0, 0, 1) and above the xy-plane.

That will be ∫∫zdxdy\displaystyle \int\int z dxdy. That cone cuts the xy-plane at z= 0 so in the circle x2+y2=1\displaystyle x^2+ y^2= 1. One way to integrate that would be to take x from -1 to 1 and, for each x, y from −1−x2\displaystyle -\sqrt{1- x^2} to 1−x2\displaystyle \sqrt{1- x^2}.

That would be the integral ∫−11∫−1−x21−x21−x2+y2dydx\displaystyle \int_{-1}^1 \int_{-\sqrt{1- x^2}}^{\sqrt{1- x^2}} 1- \sqrt{x^2+ y^2} dy dx.

Personally, because of the symmetry, I would be inclined to use polar coordinates, taking r from 0 to 1 and θ\displaystyle \theta from 0 to 2π\displaystyle 2\pi. Then the integral will be
∫02π∫01(1−r)drdθ\displaystyle \int_0^{2\pi}\int_0^1 (1- r) dr d\theta.
 
HallsofIvy said:
If f(x)=z=1−x2+y2\displaystyle f(x)= z= 1- \sqrt{x^2+y^2} then z−1=−x2+y2\displaystyle z- 1=-\sqrt{x^2+ y^2} and (z−1)2=x2+y2\displaystyle (z- 1)^2= x^2+ y^2. That is a cone with center at (0, 0, 1). You are asked to find the volume of the nappe below (0, 0, 1) and above the xy-plane.

That will be ∫∫zdxdy\displaystyle \int\int z dxdy. That cone cuts the xy-plane at z= 0 so in the circle x2+y2=1\displaystyle x^2+ y^2= 1. One way to integrate that would be to take x from -1 to 1 and, for each x, y from −1−x2\displaystyle -\sqrt{1- x^2} to 1−x2\displaystyle \sqrt{1- x^2}.

That would be the integral ∫−11∫−1−x21−x21−x2+y2dydx\displaystyle \int_{-1}^1 \int_{-\sqrt{1- x^2}}^{\sqrt{1- x^2}} 1- \sqrt{x^2+ y^2} dy dx.

Personally, because of the symmetry, I would be inclined to use polar coordinates, taking r from 0 to 1 and θ\displaystyle \theta from 0 to 2π\displaystyle 2\pi. Then the integral will be
∫02π∫01(1−r)drdθ\displaystyle \int_0^{2\pi}\int_0^1 (1- r) dr d\theta.
Click to expand...
How would I do this if the square root was actually x^2 + 2y^2 instead of x^2 + y^2, would polar coordinates still work?
 
hxnoka said:
How would I do this if the square root was actually x^2 + 2y^2 instead of x^2 + y^2, would polar coordinates still work?
Click to expand...
You could try! x2+2y2=x2+y2+y2\displaystyle x^2+ 2y^2= x^2+ y^2+ y^2 and in polar coordinates would be r2+r2sin2(θ)=r2(1+sin2(θ))\displaystyle r^2+ r^2sin^2(\theta)= r^2(1+ sin^2(\theta))
 
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