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Mth 277
Your 'question form' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
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CH 12 Practice Test Q 7
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7. Use a double integral to find the area of the region described by 0 <= r <= 3 sin (2 theta).
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I was not given any feedback on this question. So I was wondering if I completed it successfully the original time. I also watched the video to help me on the CDs given. I know polar coordinates are (r, theta) and therefore the double integral is also dr d(theta) just like it would be dydx. The bounds for r are given in the problem. The reason I used the bounds to be 0 to pi for the theta is because when we graph the 3sin(theta) given, we can see that one full wave is from 0 to pi. I was unsure exactly what to use as the function. Is the 1 correct? If either the bounds or the function is incorrect, please feel free to correct me. I really just want to make sure I understand the problem. I can easily solve the double integral from there.
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I also wanted to let you know that I plan to study tonight (07/29) and tomorrow and take the test on Thursday night. Therefore, I would like to have all questions finalized and the answers to all practice problems by mid-day tomorrow so that I can study with some worthy material. Thanks so much for your assistance :)
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r dr dTheta is the area increment. Just integrating this increment over the region is sufficient.
This is completely equivalent to integrating 1 over the region.
The correct integral would be
integral(integral(r dr, r from 0 to 3 sin(2 theta)) dTheta, theta from 0 to 2 pi).
It would also be acceptable to express this with the integration of 1 explicitly included as
integral(integral( (1) * r dr, r from 0 to 3 sin(2 theta)) dTheta, theta from 0 to 2 pi).
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Mth 277
Your 'question form' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
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Ch 12 Practice test Question 8
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8. Convert the equation 3x^2 + 3y^2 + 3z^2 = 1 to cylindrical coordinates.
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I have the equations for cylindrical coordinates. The r = sqrt(x^2+y^2) theta = arctan (y/x) and z = z. I believe I understand how to convert the equation given into this but there was no feedback given on the problem from the practice test so I wanted to reassure myself before taking the actual test. I guess the most trouble I am having is how exactly to use the equations. When I solved for x, y, and z in the original problem, I got that x is equal to +- 1/sqrt(3) y=0 and z=0. Therefore, when we plug these into the equations for the cylindrical coordinates we get r=sqrt(+-1/sqrt(3)) which the decimal approximation of that is 0.76 and I got theta and z to both be 0. Is this correct? If not, please correct it so that I can see my mistakes and study correctly for the test I plan to take Thurs (07/31) Thanks!!
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3 x^2 + 3 y^2 = 3 ( x^2 + y^2) = 3 r^2.
So the equation is 3 r^2 + 3 z^2 = 1.
You could also write this as
r^2 + z^2 = 1/3
or even
sqrt(r^2 + z^2) = sqrt(3) / 3.
You would not approximate sqrt(3).
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#$&*
Mth 277
Your 'question form' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
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Ch 12 Question 9
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9. Find the surface area of the portion of the surface z = x^2 that lies above the triangular region in the plane with vertices (0,0,0), (1,0,0) and (0,1,0).
I also used the video and the query practice problem to help me better understand this problem. I just wanted to assure that I am doing it right and could successfully do it on a test.
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I understand that according to the vertices, the surface would be in the xy plane. It would also be bounded by x and y and the line z=1-x. Therefore, we have the boundaries that x is from 0 to 1 and y is from 0 to 1-x so we are going to do dydx of the function obviously. Continuing on, we integrate the square root of (1+fx^2+fz^2) over the region. f(x,z) is equal to 2x. Therefore the sqaure root of (1+fx^2+fz^2)=sqrt(1+4x^2). The double integral would then be (integral from 0 to 1(integral from 0 to 1-x(sqrt(1+4x^2)) dydx). Did I set this up correctly or not? The best way that I learn is from knowing how to do the problem correctly. If I can practice the problem and check it over and over then I learn the steps and understand the process easier. That is just how I learn. If you could please give me the right steps then it would help a lot in my studying and preparing to take the test Thursday night. Thanks again!
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The surface z = x^2 is not in the xy plane. Anything with a nonzero z coordinate is outside that plane.
You are correct on the bounds for x and y.
fx stands for the derivative of f with respect to x. z = f(x,y) so the notation fz would have no meaning.
fx is equal to 2x, so the area factor would indeed be sqrt(1 + 4 x^2).
Your integral is correct.
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Of course you also need to perform the integration.
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