anonymous
  • anonymous
evaluate the double integral by reversing the order of operation:
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
\[\int\limits_{0}^{\sqrt{\pi}}\int\limits_{y}^{\sqrt{\pi}}Cos(x ^{2})dxdy\]
anonymous
  • anonymous
i keep getting 0
TuringTest
  • TuringTest
\[y\le x\le\sqrt\pi\]\[0\le y\le\sqrt\pi\]|dw:1332356768545:dw|

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TuringTest
  • TuringTest
or is D on the other side...
anonymous
  • anonymous
i don't know, all i was given is what i wrote
TuringTest
  • TuringTest
No, I just need to figure it out...
anonymous
  • anonymous
okay, can the answer to a double integral be a negative number?
TuringTest
  • TuringTest
why not?
anonymous
  • anonymous
idk, i thought it might be volume or something
TuringTest
  • TuringTest
not, you are integrating a function over an area if the function is negative over that area, then the integral will be negative no magic there
phi
  • phi
btw, 0 looks ok as an answer
TuringTest
  • TuringTest
yeah, it is the region I drew with the D|dw:1332357396423:dw| so\[y\le x\le\sqrt\pi\]\[0\le y\le\sqrt\pi\]implies\[x\le y\le\sqrt\pi\]\[0\le x\le\sqrt\pi\]so the integral is\[\large \int_{0}^{\sqrt\pi}\int_{x}^{\sqrt\pi}\cos(x^2)dydx\]um... did I mess up yet?
phi
  • phi
looks good
anonymous
  • anonymous
0 is the answer?
TuringTest
  • TuringTest
but you wind up with\[\int_{0}^{\sqrt\pi}\sqrt\pi\cos(x^2)-x\cos(x^2)dx\]oh wait, we can do this, I was thinking that first part can't be integrated...
TuringTest
  • TuringTest
can we do that? we're not integrating over a rectangle though
phi
  • phi
oops, I thought the limits on y are 0 to x
TuringTest
  • TuringTest
I'm a little stuck at\[\int_{0}^{\sqrt\pi}(\sqrt\pi-x)\cos(x^2)dx\]let's see....
phi
  • phi
\[\large \int\limits\limits_{0}^{\sqrt\pi}\cos(x^2)dx\int\limits\limits_{0}^{x}dy\]
TuringTest
  • TuringTest
but it's not a rectangle, it's a triangle, doesn't that mean we can't do that?
anonymous
  • anonymous
this is a confusing problem... i'm kinda lost
TuringTest
  • TuringTest
I don't think you can do that @phi I think the area has to be a rectangle with constant bounds to use that trick are you sure it applies here?
TuringTest
  • TuringTest
@allyfranken are you sure the bounds of your original problem are correct? please double-check
anonymous
  • anonymous
yup it's correct
anonymous
  • anonymous
just triple checked and it is
phi
  • phi
I believe we are integrating the lower half of the rectangle when x is the inner integral the limit is x= y to sqrt(pi) when y is the inner integral the limit is y= 0 to x
TuringTest
  • TuringTest
well even if that will work here, what are you going to do about integral cos(x^2) ?
phi
  • phi
with y= 0 to x, the inner integral gives us x we are left with cos(x^2) x dx
phi
  • phi
btw, I checked wolfram. It doesn't give how to do it, but it comes up with zero.
TuringTest
  • TuringTest
okay I got confused too, but I get it now I had the bounds wrong thanks phi, you win again :)

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