Impossible double integral?

  • Thread starter mabauti
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  • #1
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is there a closed form solution for this double integral?

[itex]\int^{2}_{1}[/itex][itex]\int^{3}_{4}[/itex][itex]\sqrt{1+4x^{2}+4y^{2}}dydx[/itex]
 

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  • #2
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is there a closed form solution for this double integral?

[itex]\int^{2}_{1}[/itex][itex]\int^{3}_{4}[/itex][itex]\sqrt{1+4x^{2}+4y^{2}}dydx[/itex]
Have you tried polar coordinates?
 
  • #3
HallsofIvy
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Exactly what do you mean by "closed form solution"? Since a limits of integration are numbers, the integral is, of course, a single number, which is about as "closed form" as you can get! But I expect that you are asking about a method of finding that number, which is a different matter.

As jedishrfu said, seeing that [itex]x^2+ y^2[/itex], I would try polar coordinates. Of course, the fact that the region of integration is a rectangle rather than a disk complicates things!

As the "[itex]\theta[/itex]" line sweeps up from the x-axis ([itex]\theta= 0[/itex]) it first touches the rectangle at (2, 3) (slope 3/2), then at the point (2, 4) (slope 2), then (1, 3) (slope 3), and finally at (1, 4) (slope 4). That gives three intervals, [itex]\theta= arctan(3/2)[/itex] to [itex]arctan(2)[/itex], [itex]\theta= arctan(2)[/itex] to [itex]arctan(3)[/itex], and [itex]\theta= arctan(3)[/itex] to [itex]arctan(4)[/itex] on which the ray enters and leaves the rectangle through different edges.

For [itex]\theta= arctan(3/2)[/itex] to [itex]arctan(2)[/itex], the ray enters the rectangle through the bottom, [itex]y= r sin(\theta)= 3[/itex] (so [itex]r= 3/sin(\theta)= 3 csc(\theta)[/itex]) and leaves through the right side, [itex]x= rcos(\theta)= 2[/itex] (so [itex]r= 2/cos(\theta)=2 sec(\theta)[/itex]. That integral will be
[tex]\int_{arctan(3/2)}^{arctan(2)}\int_{3csc(\theta)}^{2sec(\theta)}\sqrt{1+4r^2} rdrd\theta[/tex].

For [itex]\theta= arctan(2)[/itex] to [itex]arctan(3)[/itex], the ray enters the rectangle through the bottom and leaves through top, [itex]y= r sin(\theta)= 4[/itex] (so [itex]r= 4/sin(\theta)= 4 csc(\theta)[/itex]). That integral will be
[tex]\int_{arctan(2)}^{arctan(3)}\int_{3csc(\theta)}^{4csc(\theta)}\sqrt{1+4r^2} rdrd\theta[/tex].

Finally, for [itex]\theta= arctan(3)[/itex] to [itex]arctan(4)[/itex], the ray enters the rectangle through the left edge, [itex]x= r cos(\theta)= 1[/itex] (so [itex]r= 1/cos(\theta)= sec(\theta)[/itex]). That integral will be
[tex]\int_{arctan(3)}^{arctan(4)}\int_{3sec(\theta)}^{4csc(\theta)}\sqrt{1+4r^2} rdrd\theta[/tex].

Now, can you do those integrals? Would you consider that "closed form"?
 
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  • #4
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@ jedishrfu : I tried , but I couldn't figure out the intervals for theta/r

@ HallsofIvy: it worked, I can figure out the intervals for theta, but not for r. Thanks :)
 
  • #5
pasmith
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@ jedishrfu : I tried , but I couldn't figure out the intervals for theta/r

@ HallsofIvy: it worked, I can figure out the intervals for theta, but not for r. Thanks :)
The limits of the r integral are functions of theta, so you have to do the r integral before the theta integral. And the r integral is straightforward: substitute [itex]u = 1 + 4r^2[/itex].
 
  • #6
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??

I did solve the integrals.

pasmith: could you elaborate please?
 

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