review 12.6 no4

Having trouble setting up the bounds for theta in this situation z is bounded by 0 to 1 an r is bounded 0 to sqrt(2) but we realized that because the of the bounded region it is not a simple 0 to 2pi. Theta is also cut down because of the Z being only from 0 to 1 which cuts off part of the radius. We know it is something less the pi/2 but do not know what, so what do we do to find the region in which theta is bounded? 

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I guess you mean this problem?

Between y=0 and y=sqrt(2-x^2) is the half circle in the upper half plane of radius sqrt(2)--that makes theta run from 0 to pi. z between 0 and 1 has nothing to do with it.

webwork typo

Hello Professor,
I am confused about the notation used to describe the gradient vector fields.

I see that there is a problem which has the function f(x,y,z)=4x+4y+2z and we are told to take the gradient of the function  <grad>f(x,y)=_i+_j+_k

The answer that was considered correct was based on a gradient which created a R^3 vector field.

Please help me understand, I don't know what I just did.

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I see what you are saying:  problem D has a function f(x,y,z) of three variables but <grad>f(x,y) has just two variables.  That's a typo in the code, and I think I fixed it. Reload the webwork and let me know if it changed.

BTW, thanks for getting started on this early.

Lecture Notes 4/4/16-4/8/16

Lecture Notes 4/4/16

Lecture Notes 4/6/16

Lecture Notes 4/8/16

12.6 number 6

Do i have the boundaries of z correct?  Also how exactly do i calculate boundaries for x and y?  Replace something w zero or set it equal to zero or..?

Thank You,

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Yes.  To calculate the boundaries for x&y you have to understand what is the shadow of the figure on the xy-plane.  Since E is part of three dimensional space below z=4-4(x^2+y^2)  and above the xy plane, and since the xy plane is give by the equation z=0, the intersection is given by the equation 0=4-4(x^2+y^2) or x^2+y^2=1--this is the unit circle of course. Depending on the order of your integration you have to use this formula to get the limits above and below in the middle integration, while the outside integration has the limits which allow the full extent of the circle. 

12.6 Number 4

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Why am I not getting theta correct?  I used inverse tan(-3/5) and got
-.5404 and it is saying it is incorrect.

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Well, tan^{-1}(-3/5) is indeed -0.5404 as you suggest, which is the same as 2.601 as you wrote. *BUT* (-5,3) is in the second quadrant, while the formula θ=tan^{-1}(y/x) only works for (x,y) in the 1st and 4th quadrants.  The correct formula for the second and third quadrants is instead
θ=π + tan^{-1}(y/x).

Lecture Notes 3/28/16-4/1/16

Lecture Notes 3/28/16

Lecture Notes 3/30/16

Lecture Notes 4/1/16

HW section 12.5 reopened.

Hello professor, i am here at work half panicking and trying to do my homework without getting in-trouble. Can you please extend the homework until tomorrow please. I won't get out of here until 11pm. Thank you so much! 

I'm getting frustrated with how to set up 12.5.  Is there any way you can extend the assignment until Monday when I can get some face to face time with you to figure it out?  Or with a tutor since I'm sure you'll be busy with extra credit people?

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How about Sunday night?  It has been reopened.