integration problem

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February 23rd 2008, 03:14 PM
heathrowjohnny

integration problem

1. $\int_{\bold{R}} xy \ dx \ dy, \ \ \ R = \text{triangle with vertices at} \ (1,0), (2,2) \ \text{and} \ (1,2)$ .

Is this equaled to $\int\limits_{1}^{2} \int\limits_{2x-2}^{2} xy \ dy \ dx = -\frac{7}{12}$ ?

2. $\int_{\bold{R}} e^{xy} \ dx \ dy, \ \ \ R = \{(x,y) \ | \ 0 \leq x \leq 1 + (\log y)/y \}, 2 \leq y \leq 3 \}$ .

Here I am assuming that $\log y = \ln y$ .

So $0 \leq x \leq 1 + (\log 3)/3$ and $2 \leq y \leq 3$ .

So is the integral: $\int\limits_{0}^{1 + (\log 3)/3} \int\limits_{2}^{3} e^{xy} \ dy \ dx$ ?
February 23rd 2008, 03:22 PM
galactus

The lower limit in the first one should be 2x-2.

Are you finding area?. Then it shouldn't be negative.
February 23rd 2008, 04:02 PM
Krizalid

Quote:

Originally Posted by heathrowjohnny

1. $\int_{\bold{R}} xy \ dx \ dy, \ \ \ R = \text{triangle with vertices at} \ (1,0), (2,2) \ \text{and} \ (1,2)$ .

Is this equaled to $\int\limits_{1}^{2} \int\limits_{2x-2}^{2} xy \ dy \ dx = -\frac{7}{12}$ ?

You set up the double integral correctly, but its value is $\frac{11}6.$

Quote:

Originally Posted by heathrowjohnny

2. $\int_{\bold{R}} e^{xy} \ dx \ dy, \ \ \ R = \{(x,y) \ | \ 0 \leq x \leq 1 + (\log y)/y \}, 2 \leq y \leq 3 \}$ .

As it's written, it's just

$\int_2^3 {\int_0^{1 + \frac{{\ln y}}<br /> {y}} {e^{xy} \,dx} \,dy} .$

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