Recent content by CloudTenshi

  1. C

    Integration in Polar Coordinates (double integrals)

    Thanks a bunch. I realized my mistake with the r shortly after, but I had made a careless mistake because e^0 doesn't equal 0, but rather 1. I didn't realize that till I retried the problem with the integral you provided. Where the limit sign actually is doesn't really matter right? I could...
  2. C

    Integration in Polar Coordinates (double integrals)

    Homework Statement We define the improper integral (over the entire plane R^2) I as a double integral [-inf,inf]x[-inf,inf] of e^-(x^2+y^2)dA as equal to the lim as a-> inf of the double integral under Da of e^-(x^2+y^2)dA where Da is the disk with the radius a and center at the origin...
  3. C

    Finding the Maximum and Minimum Values

    Thanks. I think it all became clear for me when I found out that I was supposed to solve for L and then for the x,y,z coordinates.
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    Finding the Maximum and Minimum Values

    I just realized I made a few typos on the last one, but after solving for the x, y, and z using the value I got for L which was the sqrt(11/2) and -sqrt(11/2). From this, I was able to get the x, y, and z values which are: x=3/(2L) = plus or minus (3/2)(sqrt(2/11)) y=3/(2L) = plus or minus...
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    Finding the Maximum and Minimum Values

    I had always thought that you were supposed to just be able to look at the equations and find lambda or what not because that's what the teacher had done, but this way actually makes sense. So I solved out the L in the equation of the sphere and got sqrt(4/22) Plugging it back into my...
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    Finding the Maximum and Minimum Values

    I am having trouble solving out my lagrange multipliers. Like if I solve it out, I think I get lambda being 3/2 leaving x and y as 1 and z as 2/3.
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    Finding the Maximum and Minimum Values

    Homework Statement Find the maximum and minimum values of 3x+3y+2z on the sphere x^2+y^2+z^2=1 . Homework Equations Use of LaGrange Multipliers (maybe?) The Attempt at a Solution I have no idea if LaGrange multipliers is the way to go or not, but I took the partial derivative...
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