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  1. #1
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    region

    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 ).
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by heathrowjohnny View Post
    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).
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  3. #3
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    Quote Originally Posted by heathrowjohnny View Post
    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} .
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Krizalid View Post
    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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