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Expert: - 10/6/2009


Question
evaluate the double integral of (1+x+cos(x^2+y^2)) dxdy
with the area of integration defined by 1<=(x^2+y^2)<=4

im having a total brain freeze, any help would be awesome

Answer

Polar
Questioner:   Olivia
Category:  Advanced Math
Private:  no
 
Subject:  double integrals
Question:  evaluate the double integral of (1+x+cos(x^2+y^2)) dxdy with the area of integration defined by 1<=(x^2+y^2)<=4 im having a total brain freeze, any help would be awesome
...........................................
Hi, Olivia,

It's double-integral season, I see.  Yes, these can be confusing.  After you have done a few hundred you'll know what to do, but you are still young.

Any time you see something like  x^2 + y^2, you'll start thinking of circles.  And when you think of circles, (and are as old as I am) you'll start thinking of Polar Coordinates.

Now your standard stuff in P.C. is to integrate:

{ {
| | f(r,t) r dr dt     << t means theta; I am reducing my typing in my old age.
} }

and, as in the last problem, you can integrate in either order. (r first or t first)

What is your f(r,t)?

(1+x+cos(x^2+y^2) = 1 + r cos t + cos(r^2)

What are the boundaries?  Your theta goes  0 to 2pi, and r from 1 to 4.

{2pi {4
|    | (1 + r cos t + cos(r^2)) r dr dt
}0   }1

or vice versa:

{4 {2pi
|  | (1 + r cos t + cos(r^2)) dt r dr
}1 }0

I think the  r dr is a nice thing to have.  Let's do the r thing first.

{4
| (1 + r cos t + cos(r^2)) r dr
}1

{4
| (r + r^2 cos t + r cos(r^2)) dr
}1

r^2/2 + r^3/3 cos t + sin(r^2)/2, from  1 to 4.

8 + 64/3 cos t + sin(16)/2 - (1/2 + 1/3 cos t + sin(1)/2)

8 + 64/3 cos t + sin(16)/2 - 1/2 - 1/3 cos t - sin(1)/2

15/2 + 21 cos t + 1/2(sin(16) - sin(1))

15/2 + 1/2(sin(16) - sin(1)) + 21 cos t   << rearranged.

Now integrate that for t going from 0 to 2pi.  I think the cos t part integrates to 0, so all you need is the constant part times 2pi.

I think the other way works, too.

Paul Klarreich

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