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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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I think I figured out why I was confused with polar form double integrals. I was under the impression that we had to account for the area change by including r in the integral wrt `dr, which is correct. Then when I went to attempt the Query I the first problem was Evaluate the double integral Int( Int(dr, 0, 1 + sin(theta)) dθ), 0 , pi). Which is not possible right because it would have to be Int( r dr,0,1 + sin(`theta)) = (1+sin(`theta))^2/2???

@& That integral is perfectly legitimate. It's simply a double integral; there's no mention of it representing any sort of area calculation.

That integral could arise from a plane region. For example, if the density is proportional to 1 / r, the mass calculation could look something ike this.*@

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????I'm sending the actual question from my access page with the comments????

Question: `q001. Evaluate the double integral Int( Int(dr, 0, 1 + sin(theta)) dθ), 0 , pi). Sketch and describe the region of integration.In standard notation the integral is

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Your solution:

Int( 1 dr) = r, evaluated at limits

[1+sin(`theta)] - 0 = 1+sin(`theta)

Int(1+sin(`theta) d`theta) = Int(1 d`theta) + Int(sin(`theta) d`theta)

=`theta - cos(`theta), evaluated at limits

[`pi - cos(`pi)] - [0 - cos(0)]

=`pi + 2

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Given Solution:

An antiderivative of 1 is r. The change in the antiderivative between r = 0 and r = 1 + sin(theta) is 1 + sin(theta).

Integrating this result with respect to theta we get antiderivative theta - cos(theta).

The change in this quantity between theta = 0 and theta = pi is

pi - cos(pi) - (0 - cos(0) ) = pi - (-1) - (-1) = pi + 2.

Additional note:

Between theta = 0 and theta = pi the value of sin(theta) goes from 0 to 1 and back to 0, so the value of 1 + sin(theta) goes from 1 to 2 to 1.

The graph of 1 + sin(theta) from theta value 0 to 2 pi is given below, and the portion of the graph from theta values 0 to pi is in light blue, the portion above the x axis. It should be clear that this region contains a circle of radius 1 (visualize the circle centered at (0, 1)) and therefore has somewhat more area. The area of the circle would be pi, so the result pi + 2 makes good sense.

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Self-critique (if necessary):

Still can’t see the graphs the best but getting better. So we only have the form Int(Int( “”” r dr) d`theta) when converting from x,y values to polar coord right?

@& Nothing is being converted to polar coordinates. This integral was in polar coordinates to start with. We could thinkg about what the corresponding integral would be in rectangular coordinates, but let's not get into that on this problem.*@

I mean that it is not necessary for our integral that is wrt `dr to have a mult of r in the integral, unless we are converting to polar coord and if not why does integral wrt `dr not have mult of r in the integral??????????

@& The area increment in polar coordinates is r dr dtheta. So this integral represents the area within the curve r = 1 + theta, for theta = 0 to pi. Just be sure you understand the graphs.*@

@& Actually this comment is inaccuate. The area increment is r dr dTheta, and that r isn't part of the given integral. See my previous note for a possible interpretation.

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