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How to evaluate int 2x3y dA using change of variablesby AeroFunk
Tags: jacobians 
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#1
Dec204, 01:33 AM

P: 40

Let R be the region bounded by the graphs of x+y=1, x+y=2, 2x3y=2, and 2x3y+5. Use the change of variables:
[tex] x=1/5(3u+v)[/tex] [tex]y=1/5(2uv)[/tex] to evaluate the integral: [tex] \iint(2x3y)\,dA [/tex] I found the jachobian to be 1/5 and the limits of integration to be 1<=u<=2 2<=v<=5 so i set up the integral like this: [tex] \frac{1}{5}\int_{2}^{5}\int_{2}^{1} vdv[/tex] and I get 21/5 which doesn't seem right(a negitive number??),what am I doing wrong? 


#2
Dec204, 02:15 AM

P: 290

I'm getting 21/10, but that's just because you forgot a 1/2 in the integration. I am wondering, though, why your bounds of integration on u are going from 2 to 1 instead of 1 to 2, especially since you don't evalute the integral as such.
What's wrong with a negative number? The integral in uv obviously isn't going to be negative, and the Jacobian is in fact negative, so obviously the answer should be negative. From the xy integral, we're integrating 2x3y. If y is sufficiently large compared to x, the integral will be negative, as in this case. J 


#3
Dec204, 02:31 AM

P: 40

Oh sorry the 2 to 1 is just a typo , and lol yea it would be negitive (have been doing to many volume problems lately)
thanks for the help 


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