Solving Integration Problem with Double Integral in Polar Coordinates

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Homework Help Overview

The discussion revolves around a double integral involving a function defined in polar coordinates, specifically integrating over a region where \(x^2 + y^2 \geq 2\). The original poster expresses difficulty in calculating the integral using Cartesian coordinates and seeks guidance on transitioning to polar coordinates.

Discussion Character

  • Exploratory, Problem interpretation, Assumption checking

Approaches and Questions Raised

  • Participants discuss the nature of the integral and the region of integration, with one questioning whether the region is infinite. Others clarify the integral's setup and express the need to understand the geometric context of the problem.

Discussion Status

The conversation is ongoing, with participants exploring different interpretations of the problem and the implications of the defined region. Some have provided clarifications on the integral's formulation, while others are attempting to connect the integral to a broader context involving a vector field and surface integrals.

Contextual Notes

There are references to a surface defined by a sphere and the implications of the vector field involved, indicating that the problem may have multiple layers of complexity. The original poster's constraints and the nature of the problem suggest that further exploration of assumptions and definitions is necessary.

JaysFan31
I need to integrate
double integral 2(sqrt(9-x^2-y^2)(-x^2+y^2-2)dxdy with the bounds determined by the fact that x^2+y^2 is greater than or equal to 2.

This integral is impossible to calculate using cartesian coordinates. How would I do it using polar coordinates?
 
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[tex]\int \!\!\! \int 2\sqrt{(9-x^2-y^2)(-x^2+y^2-2)}\,dx\,dy[/tex]

[tex]x^2 + y^2 \geq 2[/tex]

Is that the integral in question?
 
That's an infinite region isn't it?
 
Yes that's it, the one cepheid wrote.

Well it's obviously part of a problem so here goes:
Let S be the subset of the surface of the sphere x^2+y^2+z^2=9 for which x^2+y^2 is greater than or equal to 2. Let F be the vector field defined by (-y, x, xyz).

Computer double integral(curlF)*ndS where * is the dot product and S is oriented so that the unit normal n to S points away from the enclosed volume.

I have the vector field= -yi+xj+xjzk.
curlF (after calculation)=xzi-yzj+2k
G(x,y,z)=9-z^2-x^2-y^2 and gradG=-2xi-2yj-2zk
Thus, dS=(-2xi-2yj-2zk)dxdy
The dot product of curlF and dS=-2(x^2)z+2(y^2)z-4z
I plug in for z=sqrt(9-x^2-y^2) and I'm in this predicament.
 
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