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June 20th, 2018, 11:22 AM   #1
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Analysis II Exam - Doubts

Hi everyone, sorry to post a question right after registration but I have been having difficulty with this problem. I hope someone could help me.

I had these two questions in an exam and I still am not sure about the answers.

First one: Calculate the mass of the solid with vertices (1,1,0), (1,-1,0), (-1,-1,0), (-1,1,0), (0,0,2), where the density is given by f(x,y,z) = |x|.

And the second question is just a simple one about changing the order of integration, but I am having second doubts...



Hope you can help. Thank you!

Last edited by skipjack; June 20th, 2018 at 12:21 PM.
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June 20th, 2018, 12:22 PM   #2
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Can you post your attempts?
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June 20th, 2018, 01:57 PM   #3
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For the second problem graph the domain of integration - it will be a triangle. Changing the order of integration becomes obvious.
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June 21st, 2018, 02:46 AM   #4
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Quote:
Originally Posted by skipjack View Post
Can you post your attempts?
For the domain of integration, I was able to find this value and hope it is right.


For the first question I don't really have a solid attempt, everything I tried wasn't working.

I basically tried to follow this thread by changing the density value which in my case is f(x,y,z)= |x|.
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June 21st, 2018, 05:57 AM   #5
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Ok, I discovered my mistake for the second question.
I have posted how I tried to resolve the first question but it has to be approved for it to be posted. In the meantime if someone could help me with the first question I would appreciate it.
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June 21st, 2018, 12:12 PM   #6
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I think you tried to link to an image from W|A that no longer exists.
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July 1st, 2018, 05:25 AM   #7
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For the second problem start by graphing the area of integration. The "outer" integral is for y from 0 to 5 so draw horizontal lines at y= 0 (the x-axis) to y= 5. The "inner" integral is for x from 0 to y/2- x/2 so draw the vertical line x= 0 (the y-axis) and x= y/2- 1/2 which is the same as y= 2x+ 1. The lines y= 2x+ 1 and y= 5 intersect where y= 2x+ 1= 5 so at x= 2, y= 5. The area of integration is the triangle with vertices at (0, 1), (0, 5), and (2, 5).

To cover the same region, x goes from 0 up to 2. And, for each x, y goes from the line y= 2x+ 1 to y= 5.
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July 1st, 2018, 05:43 AM   #8
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The first problem involves the pyramid with base the square in the xy-plane with vertices (1, 1, 0), (1, -1, 0), (-1, -1, 0), (-1, 1, 0), and tip at (0, 0, 2).

One edge of that pyramid is the line from (1, 1, 0) to (0, 0, 2) and it is easy to see that we can write that with parametric equation x= 1- t, y= 1- t. z= 2t with t going from 0 to 1.

Another edge is the line from (1, -1, 0) to (0, 0, 2). That has parametric equations x= 1- t, y= t- 1, z= 2t with t going from 0 to 1.

The third edge is the line from (-1, -1, 0) to (0, 0, 2). That has parametric equations x= t- 1, y= t- 1, z= 2t with t going from 0 to 1.

Finally, the fourth edge is the line from (-1, 1, 0) to (0, 0, 2). That has parametric equations x= t- 1, y= 1- t, z= 2t.

Since the "density" function is |x|, depending on x only, imagine cutting that pyramid with cross sections at a fixed value of x.

Such a cross section is a rectangle with vertices at (x, x, 0), (x, -x, 0), (x, x, 2(x+1)), and (x, -x, 2(x+ 1)). That has area $(2x)(2(x+ 1))= 4x^2+ 4x$. Taking the thickness to be "dx" it has volume $(4x^2+ 4x)dx$ and mass $|x|(4x^2+ 4x)dx$. For x< 0 that is $(-4x^3- 4x^2)dx$ and for x> 0 it is $(4x^3+ 4x^2)dx$. The total mass is $-4\int_{-1}^0 (x^3+ x^2) dx+ 4\int_0^1 (x^3+ x^2) dx$.

Last edited by Country Boy; July 1st, 2018 at 05:47 AM.
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