Evaluate Double integral

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tony873004
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Homework Statement


Evaluate the integral [tex]\int {\int\limits_R {\left( {x + y} \right)\,dA} } [/tex] where R is the region that lies to the left of the y-axis between the circles [tex]x^2 + y^2 = 1[/tex] and [tex]x^2 + y^2 = 4[/tex] by changing to polar coordinates.



Homework Equations


x=r cos theta
y=r sin theta


The Attempt at a Solution


my effort:

[tex]\begin{array}{l}
r_{inner} = \sqrt 1 = 1,\,\,r_{outer} = \sqrt 4 = 2 \\
R = \left\{ {\left( {r,\theta |1 \le r \le 4,\,\frac{{3\pi }}{2}\,\theta \le \pi } \right)} \right\} \\
x = r\cos \left( \theta \right),\,\,\,y = r\sin \left( \theta \right) \\
\int\limits_1^2 {\int\limits_{\pi /2}^{3\pi /2} {\left( {r\cos \left( \theta \right) + r\sin \left( \theta \right)} \right)\,d\theta \,dr} } \\
\end{array}[/tex]

The solution shows an extra instance of r in the integral. If the original question is for (x+y) then why isn't it simply r cos + r sin, rather than r(r cos + r sin)?

Homework Statement





Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #2
378
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Assuming everything else is correct:

there is an additional r because of Jacobian
you will learn co-od transformations soon
when you transform, area changes, and that
r takes into account of that
 
  • #3
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the area element in polar coordinates is :

[tex] dA = rdrd\theta[/tex]
that is where the extra r comes from.
 
  • #4
tony873004
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Thanks!

I was about to say never mind because I just found the formula in the book. It gives the extra r.

But I'm glad I asked, because now I understand where it comes from.
 
  • #5
208
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indeed, you can see this also if you draw out the picture of the infinitesimal pieces: rdtheta will be a small slice of length. multiplying by dr will give you approximately a small square. so dxdy =~ rdrdtheta
 

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