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## Find the flux of F over Q where Q is bounded?

Find the flux of F over Q where Q is bounded by z=sqrt(x^2+y^2), x^2+y^2=1 and z=0, F=<y^2, x^2*z, z^2>. (Answer: pi/2)

The divergence of the vector field is 2z, which is what I've found. And I know that x^2+y^2=r^2, z=r, r=0, x^2+y^2=0, how should I find the points of the integral and set up for the integral?

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# 1 Answer

Hi Sun,

You're on the right path. To calculate this flux integral, you should first realize that parametrizing the surface Q would be hard, so it's better to try to use the Divergence Theorem and instead compute the integral of the divergence of F not over Q but over the volume enclosed by Q. You're right that div F = 2z. So you just need to set up the volume integral.

Note that x2+y2=1 determines a cylinder of radius 1. So the volume is best described in cylindrical coordinates, which is just polar coordinates in the xy-plane together with the z coordinate. If you sketch the surface Q you should be able to convince yourself that the volume V will be parametrized by theta in [0,2pi), z in [0,1], and r in [z,1]. (To see that last part, notice that z=sqrt(x2+y2). But x2+y2=r2 in polar coordinates, so this says z=r.)

These are your limits of integration, so you just carry out the integral of 2z in cylindrical coordinates with the cylindrical volume element r dr dtheta dz. If you do this, you do indeed get an answer of pi/2. Let me know if you need more help.

Best wishes,

Matt

B.S. Mathematics, MIT