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Problem 23 Easy Difficulty

Evaluate the appropriate determinant to show that the Jacobian
of the transformation from Cartesian $\rho \phi \theta$ -space to Cartesian $x y z-$
space is $\rho^{2} \sin \phi$.

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Video Transcript

{'transcript': "were asked to find the Jacoby in of the transformation from Cartesian Row Phi data space to Cartesian X y Z space. Well, what this is really is a question is about spherical coordinates. So our parametric equations are X equals rose sign fee Cosine theta. Why equals Rose sine phi Sign of Theta and Z equals row Co sign Phi and therefore Jacoby in of this transformation the X y Z, dear Oh, bye. Data is the determinant of the three by three matrix, which is while the partial of X with respect to row is signed. Fi cosine theta partial of light with respect to row is signed fi No, sorry. That's mistake. Partial of X with respect to fi is row cosine Phi CoSine data in the partial of X with respect to theta is negative Rose signed fi signed data partial of why this is also going to be signed fi sign of data row Cosine Phi Sign of data and row sign of fi, cosine theta And finally, the partial of Z with respect to row is cosine Phi Partial of Z with respect to fi is negative Rose sine phi in the partial of Z with respect to data is zero. So to evaluate, this tournament will expand across the bottom row. And so this is equal to cosign five times the determinant of the upper two by two Matrix row Cosine Phi cosine theta negative Rose sine phi sine data and row Cosine phi signed data Rose sine phi cosine data. And then we have minus negative Rose sine phi or plus rose sine phi times the determinant of the matrix sine phi cosine data negative. Rose signed Fi signed data sci fi signed data and Rose sine phi cosine data in evaluating we get to cosign five times. In fact, er out a row squared. Here we have a row squared cosign five times and then we have cosign Phi sine phi cosine squared data plus cosine phi sine phi sine squared data, which is just sign Phi cosine phi. So we get cosine squared. If I sign Phi in total Plus and then we have ro times again. We can factor at a rose. So we get rose squared sign five times. Then this is if sine phi squared sine squared fi oh sine squared data plus sine squared phi sine squared data to factor out the sine squared fi ve it just another sine squared fives that becomes Sign Cubed Fi And then factoring out the road squared and factoring out a sign. Five we get row squared sign five times cosine squared five plus sine squared five, which is simply Rose squared sine phi. And this is, of course, the same answer."}

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