Recent content by letalea

  1. L

    Double Integral (underneath a surface and above a square)

    Yes, that makes perfect sense. Thank you!
  2. L

    Double Integral (underneath a surface and above a square)

    Could you explain this mistake to me? As I am not seeing how I come to get the coefficients. But yes you are right, when you factor in the coefficients it does give me one of the possible answers the prof gave us!
  3. L

    Double Integral (underneath a surface and above a square)

    Homework Statement The volume underneath the surface z= y/ (1+xy) and above the square {(x,y)| 2≤x≤3 , 3≤ y≤ 4} is:Homework Equations Please see attachment. The Attempt at a Solution Please see attachment for solution. My professor had provided us with 8 possible solutions (where only one...
  4. L

    Finding the Voltage Rating of a Battery in Fig. 3

    Homework Statement The field generated in Fig. 3 is due to a battery connected between the center conductor and the sheath. What is the voltage rating of the battery? (i.e. what is the electric potential of the positive terminal with respect to the negative terminal of the battery?) I...
  5. L

    Multivariable calculus. The chain rule.

    Thank you I will give that a try!
  6. L

    Multivariable calculus. The chain rule.

    Okay I hope this is what you were wanting me to answer... \partialz/\partialu = (\partialz/\partialx) (\partialx/\partialu) + (\partialz/\partialy)(\partialy/\partialu) =(2xysin(u)tan(v))(\partialx/\partialu) +(x^2sin(u)tan(v))(\partialy/\partialu) \partialz/\partialv=...
  7. L

    Multivariable calculus. The chain rule.

    Yes sorry, it was a posting mistake, the question was z=x^2ysin(u)tan(v). Can you see any errors in how I solved the partial derivatives, as I have tried multiple times and can not seem to find the correct solution.
  8. L

    Multivariable calculus. The chain rule.

    Homework Statement Let x=x^2ysin(u)tan(v), where x(u,v) and y(u,v) are smooth functions that, when evaluated at u=1 and v=-3 satisfy x=2.112, y=4.797, \partialx/\partialu = -3.491, \partialx/\partialv = -2.230 , \partialy/\partialu = 1.787 , \partialy/\partialv = 1.554. Then the...
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