region

**heathrowjohnny** · February 23rd 2008, 06:01 PM

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

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

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

**Jhevon** · February 23rd 2008, 06:06 PM

Originally Posted by heathrowjohnny

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

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

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

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).

**Krizalid** · February 23rd 2008, 06:15 PM

Originally Posted by heathrowjohnny

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

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

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

**Jhevon** · February 23rd 2008, 06:16 PM

Originally Posted by Krizalid

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

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

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

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