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Calc 3 Double Integrals Problem
- Thread starter hxnoka
- Start date
D
Deleted member 4993
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What are the methods that you have been taught?View attachment 23795
(I am mainly stuck on how to write the integral given this information, but would appreciate a fully solved solution too.)
Thank you.
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.What are the methods that you have been taught?
D
Deleted member 4993
Guest
make an approximate sketch of f(x,y) and show us in a picture.View attachment 23795
(I am mainly stuck on how to write the integral given this information, but would appreciate a fully solved solution too.)
Thank you.
HallsofIvy
Elite Member
- Joined
- Jan 27, 2012
- Messages
- 7,763
If \(\displaystyle f(x)= z= 1- \sqrt{x^2+y^2}\) then \(\displaystyle z- 1=-\sqrt{x^2+ y^2}\) and \(\displaystyle (z- 1)^2= x^2+ y^2\). 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 \(\displaystyle \int\int z dxdy\). That cone cuts the xy-plane at z= 0 so in the circle \(\displaystyle x^2+ y^2= 1\). One way to integrate that would be to take x from -1 to 1 and, for each x, y from \(\displaystyle -\sqrt{1- x^2}\) to \(\displaystyle \sqrt{1- x^2}\).
That would be the integral \(\displaystyle \int_{-1}^1 \int_{-\sqrt{1- x^2}}^{\sqrt{1- x^2}} 1- \sqrt{x^2+ y^2} dy dx\).
Personally, because of the symmetry, I would be inclined to use polar coordinates, taking r from 0 to 1 and \(\displaystyle \theta\) from 0 to \(\displaystyle 2\pi\). Then the integral will be
\(\displaystyle \int_0^{2\pi}\int_0^1 (1- r) dr d\theta\).
That will be \(\displaystyle \int\int z dxdy\). That cone cuts the xy-plane at z= 0 so in the circle \(\displaystyle x^2+ y^2= 1\). One way to integrate that would be to take x from -1 to 1 and, for each x, y from \(\displaystyle -\sqrt{1- x^2}\) to \(\displaystyle \sqrt{1- x^2}\).
That would be the integral \(\displaystyle \int_{-1}^1 \int_{-\sqrt{1- x^2}}^{\sqrt{1- x^2}} 1- \sqrt{x^2+ y^2} dy dx\).
Personally, because of the symmetry, I would be inclined to use polar coordinates, taking r from 0 to 1 and \(\displaystyle \theta\) from 0 to \(\displaystyle 2\pi\). Then the integral will be
\(\displaystyle \int_0^{2\pi}\int_0^1 (1- r) dr d\theta\).
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?If \(\displaystyle f(x)= z= 1- \sqrt{x^2+y^2}\) then \(\displaystyle z- 1=-\sqrt{x^2+ y^2}\) and \(\displaystyle (z- 1)^2= x^2+ y^2\). 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 \(\displaystyle \int\int z dxdy\). That cone cuts the xy-plane at z= 0 so in the circle \(\displaystyle x^2+ y^2= 1\). One way to integrate that would be to take x from -1 to 1 and, for each x, y from \(\displaystyle -\sqrt{1- x^2}\) to \(\displaystyle \sqrt{1- x^2}\).
That would be the integral \(\displaystyle \int_{-1}^1 \int_{-\sqrt{1- x^2}}^{\sqrt{1- x^2}} 1- \sqrt{x^2+ y^2} dy dx\).
Personally, because of the symmetry, I would be inclined to use polar coordinates, taking r from 0 to 1 and \(\displaystyle \theta\) from 0 to \(\displaystyle 2\pi\). Then the integral will be
\(\displaystyle \int_0^{2\pi}\int_0^1 (1- r) dr d\theta\).
D
Deleted member 4993
Guest
Excellent question!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?
Try it and tell us what you found......
HallsofIvy
Elite Member
- Joined
- Jan 27, 2012
- Messages
- 7,763
You could try! \(\displaystyle x^2+ 2y^2= x^2+ y^2+ y^2\) and in polar coordinates would be \(\displaystyle r^2+ r^2sin^2(\theta)= r^2(1+ sin^2(\theta))\)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?