# region

• Feb 23rd 2008, 06:01 PM
heathrowjohnny
region
What is the region $\displaystyle R$ over which you integrate when evaluating the double integral $\displaystyle \int_{1}^{2} \int_{1}^{x} \frac{x}{\sqrt{x^2+y^2}} \ dy \ dx$? Rewrite this as an iterated integral first with respect to $\displaystyle x$, then with respect to $\displaystyle y$. Evaluate this integral. Which order of integration is easier?

Is the region simply $\displaystyle R = \{(x,y) \ | \ 1 \leq x \leq 2, \ 1 \leq y \leq x \}$?

Rewriting it, I get $\displaystyle \int_{1}^{x} \int_{1}^{2} \frac{x}{\sqrt{x^2+y^2}} \ dx \ dy$. And evaluating it should be straightforward (guessing that it will be easier to do it first with respect to $\displaystyle x$).
• Feb 23rd 2008, 06:06 PM
Jhevon
Quote:

Originally Posted by heathrowjohnny
What is the region $\displaystyle R$ over which you integrate when evaluating the double integral $\displaystyle \int_{1}^{2} \int_{1}^{x} \frac{x}{\sqrt{x^2+y^2}} \ dy \ dx$? Rewrite this as an iterated integral first with respect to $\displaystyle x$, then with respect to $\displaystyle y$. Evaluate this integral. Which order of integration is easier?

Is the region simply $\displaystyle R = \{(x,y) \ | \ 1 \leq x \leq 2, \ 1 \leq y \leq x \}$?

Rewriting it, I get $\displaystyle \int_{1}^{x} \int_{1}^{2} \frac{x}{\sqrt{x^2+y^2}} \ dx \ dy$. And evaluating it should be straightforward (guessing that it will be easier to do it first with respect to $\displaystyle x$).

you are correct.

going with x first, you can do a simple substitution. doing y first you need a trig substitution (or Krizalid can find another way, but it will be more work, i think).
• Feb 23rd 2008, 06:15 PM
Krizalid
Quote:

Originally Posted by heathrowjohnny
Rewriting it, I get $\displaystyle \int_{1}^{x} \int_{1}^{2} \frac{x}{\sqrt{x^2+y^2}} \ dx \ dy$. And evaluating it should be straightforward (guessing that it will be easier to do it first with respect to $\displaystyle x$).

No, it looks like you tried to apply Fubini-type Theorem here.

Having $\displaystyle \int_1^2 {\int_1^x {\frac{x} {{\sqrt {x^2 + y^2 } }}\,dy\,dx} },$ it's obviously easier to compute the integral in the $\displaystyle dx\,dy$ order, which is $\displaystyle \int_1^2 {\int_y^2 {\frac{x} {{\sqrt {x^2 + y^2 } }}\,dx} \,dy} .$
• Feb 23rd 2008, 06:16 PM
Jhevon
Quote:

Originally Posted by Krizalid
No, it looks like you tried to apply Fubini-type Theorem here.

Having $\displaystyle \int_1^2 {\int_1^x {\frac{x} {{\sqrt {x^2 + y^2 } }}\,dy\,dx} },$ it's obviously easier to compute the integral in the $\displaystyle dx\,dy$ order, which is $\displaystyle \int_1^2 {\int_y^2 {\frac{x} {{\sqrt {x^2 + y^2 } }}\,dx} \,dy} .$

ah, i forgot to check the limits he/she has.