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    Consider forces \mathbf{F}=(3x^2yz^2,2x^3yz,x^3z  ^2) and \mathbf{G}=(3x^2y^2z,2x^3yz,x^3y  ^2).

    Compute the work done, given by the line integrals \int\mathbf{F}.d\mathbf{r} and \int\mathbf{G}.d\mathbf{r} along the paths consisting of straight line segments joining the specified points:

    i)(0,0,0)->(1,1,1)
    ii)(0,0,0)->(0,0,1)->(0,1,1)->(1,1,1)
    iii)(0,0,0)->(1,0,0)->(1,1,0)->(1,1,1)

    I got the answer to every single on to be 1, which doesn't seem right. Am I right?
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    Hmm, I get 1/3 for F (iii), but I fear I've made a mistake. G is correct since f(x,y,z) = x^3y^2z has the desired grad vector.
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    Sorry Glut, you're right, made a stupid mistake.

    In other news what are \mathbf{e_r} and \mathbf{e_{\theta}}? Or do we not care and care only for the relations between e_1 and e_2 and the aforementioned vectors (e_1=e_r cos e_{theta}, e_2=e_r sin e_{theta})?
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    (Original post by around)
    Sorry Glut, you're right, made a stupid mistake.

    In other news what are \mathbf{e_r} and \mathbf{e_{\theta}}? Or do we not care and care only for the relations between e_1 and e_2 and the aforementioned vectors (e_1=e_r cos e_{theta}, e_2=e_r sin e_{theta})?
    I'm not really sure myself, but I found some definitions of the basis vectors in polar coordinates here: http://en.wikipedia.org/wiki/Del_in_...al_coordinates (polar coordinates just being cylindrical coordinates without a z-axis).
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    (Original post by around)
    In other news what are \mathbf{e_r} and \mathbf{e_{\theta}}?
    These are unit vectors in the 'radial' and 'tangential' directions respectively, but of course they're not fixed, because where you are alters the direction they point in. Some nonsense, basically. But at each point, e_r and e_theta give you an orthonormal basis.

    It's almost certain that none of this is important for whatever your question is - do you have a specific question?
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    Basically, I'm trying to derive the formula for grad f in polar co-ordinates.
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    Done that question, trying to learn volume integrals:

    By using a suitable change of variables, calculate the volume within an ellipsoid

    \dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} + \dfrac{z^2}{c^2} \leq 1.

    First obvious thought: change into spherical co-ordinates.  x=a sin\theta sin\phi\ y= b sin \theta sin \phi\ z=c cos\theta (r is uniformly 1), as \dfrac{\partial x}{\partial r}=0 and similarly for y and z. However, this gives a determinant of the Jacobian to be 0. What have I missed?
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    (Original post by around)
    Done that question, trying to learn volume integrals:

    By using a suitable change of variables, calculate the volume within an ellipsoid

    \dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} + \dfrac{z^2}{c^2} \leq 1.

    First obvious thought: change into spherical co-ordinates.  x=a sin\theta sin\phi\ y= b sin \theta sin \phi\ z=c cos\theta (r is uniformly 1), as \dfrac{\partial x}{\partial r}=0 and similarly for y and z. However, this gives a determinant of the Jacobian to be 0. What have I missed?
    (I assume here one of x and y was meant to be (const) sin theta cos phi?)

    I'm not too sure, but I think you've misunderstood how spherical polars work. At least, you've represented three variables (x, y, z) in terms of two (theta, phi) - basic matrices/simultaneous equations work should tell you that's not going to work somehow. All of the partial derivatives wrt r are going to be zero, so the Jacobian is going to be zero. You spotted this yourself when you said "r is uniformly 1", even though to calculate the volume you're somehow going to have to integrate from r=0 (the centre) to r=1 (the boundary).

    Maybe try x/a = r sin theta cos phi, y/b = r sin theta sin phi, z/c = r cos theta.
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    Oh, right, thank you very much (yes, x was meant to be sin theta cos phi). I wish the VC lecturer would cover spherical coordinates before setting us questions on them :<
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    (Original post by around)
    Oh, right, thank you very much (yes, x was meant to be sin theta cos phi). I wish the VC lecturer would cover spherical coordinates before setting us questions on them :<
    Mmm, I know the feeling. You should try to just see spherical coords as a special case of a general change of coords, though (and engrave the various Jacobians onto your retinas the day before the exams) - there's nothing really very tricky about them, but you're right that they do feel a bit scary first time round.
 
 
 
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