Double integral - transformation












10












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I'm trying to calculate $$iint_{Omega } e^{(x+y^2)^{3/2}} ,mathrm{d}A,$$ where $$Omega ={x,y>0 : x+yleq 2}. $$
Not sure where to go with it. I need to find a transformation and then calculate the integral. Any pointers?






integration multivariable-calculus definite-integrals







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  • $begingroup$
    Are you sure the question had $x+y^2$ in the exponent and not $x^2+y^2$?
    $endgroup$
    – pre-kidney
    Apr 25 '17 at 16:33










  • $begingroup$
    Yeah it's defininitely $x+y^2$.
    $endgroup$
    – Anon
    Apr 25 '17 at 16:37










  • $begingroup$
    Okay, can you also verify that the region of integration is correct?
    $endgroup$
    – pre-kidney
    Apr 25 '17 at 16:37










  • $begingroup$
    (The reason I am asking these questions is because the integral you posted appears to have no closed form expression.)
    $endgroup$
    – pre-kidney
    Apr 25 '17 at 16:40










  • $begingroup$
    I've tried with this coordinate change, but it doesn't seems easy either $begin{cases} x=t^2-s^2\ y=s end{cases}$
    $endgroup$
    – Rafa Budría
    Apr 25 '17 at 18:14


















10












$begingroup$


I'm trying to calculate $$iint_{Omega } e^{(x+y^2)^{3/2}} ,mathrm{d}A,$$ where $$Omega ={x,y>0 : x+yleq 2}. $$
Not sure where to go with it. I need to find a transformation and then calculate the integral. Any pointers?






integration multivariable-calculus definite-integrals







share|cite|improve this question











$endgroup$












  • $begingroup$
    Are you sure the question had $x+y^2$ in the exponent and not $x^2+y^2$?
    $endgroup$
    – pre-kidney
    Apr 25 '17 at 16:33










  • $begingroup$
    Yeah it's defininitely $x+y^2$.
    $endgroup$
    – Anon
    Apr 25 '17 at 16:37










  • $begingroup$
    Okay, can you also verify that the region of integration is correct?
    $endgroup$
    – pre-kidney
    Apr 25 '17 at 16:37










  • $begingroup$
    (The reason I am asking these questions is because the integral you posted appears to have no closed form expression.)
    $endgroup$
    – pre-kidney
    Apr 25 '17 at 16:40










  • $begingroup$
    I've tried with this coordinate change, but it doesn't seems easy either $begin{cases} x=t^2-s^2\ y=s end{cases}$
    $endgroup$
    – Rafa Budría
    Apr 25 '17 at 18:14
















10












10








10


4



$begingroup$


I'm trying to calculate $$iint_{Omega } e^{(x+y^2)^{3/2}} ,mathrm{d}A,$$ where $$Omega ={x,y>0 : x+yleq 2}. $$
Not sure where to go with it. I need to find a transformation and then calculate the integral. Any pointers?






integration multivariable-calculus definite-integrals







share|cite|improve this question











$endgroup$




I'm trying to calculate $$iint_{Omega } e^{(x+y^2)^{3/2}} ,mathrm{d}A,$$ where $$Omega ={x,y>0 : x+yleq 2}. $$
Not sure where to go with it. I need to find a transformation and then calculate the integral. Any pointers?






integration multivariable-calculus definite-integrals




integration multivariable-calculus definite-integrals






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Feb 11 '18 at 1:48









Saad

19.7k92352




19.7k92352










asked Apr 25 '17 at 16:17









AnonAnon

408315




408315












  • $begingroup$
    Are you sure the question had $x+y^2$ in the exponent and not $x^2+y^2$?
    $endgroup$
    – pre-kidney
    Apr 25 '17 at 16:33










  • $begingroup$
    Yeah it's defininitely $x+y^2$.
    $endgroup$
    – Anon
    Apr 25 '17 at 16:37










  • $begingroup$
    Okay, can you also verify that the region of integration is correct?
    $endgroup$
    – pre-kidney
    Apr 25 '17 at 16:37










  • $begingroup$
    (The reason I am asking these questions is because the integral you posted appears to have no closed form expression.)
    $endgroup$
    – pre-kidney
    Apr 25 '17 at 16:40










  • $begingroup$
    I've tried with this coordinate change, but it doesn't seems easy either $begin{cases} x=t^2-s^2\ y=s end{cases}$
    $endgroup$
    – Rafa Budría
    Apr 25 '17 at 18:14




















  • $begingroup$
    Are you sure the question had $x+y^2$ in the exponent and not $x^2+y^2$?
    $endgroup$
    – pre-kidney
    Apr 25 '17 at 16:33










  • $begingroup$
    Yeah it's defininitely $x+y^2$.
    $endgroup$
    – Anon
    Apr 25 '17 at 16:37










  • $begingroup$
    Okay, can you also verify that the region of integration is correct?
    $endgroup$
    – pre-kidney
    Apr 25 '17 at 16:37










  • $begingroup$
    (The reason I am asking these questions is because the integral you posted appears to have no closed form expression.)
    $endgroup$
    – pre-kidney
    Apr 25 '17 at 16:40










  • $begingroup$
    I've tried with this coordinate change, but it doesn't seems easy either $begin{cases} x=t^2-s^2\ y=s end{cases}$
    $endgroup$
    – Rafa Budría
    Apr 25 '17 at 18:14


















$begingroup$
Are you sure the question had $x+y^2$ in the exponent and not $x^2+y^2$?
$endgroup$
– pre-kidney
Apr 25 '17 at 16:33




$begingroup$
Are you sure the question had $x+y^2$ in the exponent and not $x^2+y^2$?
$endgroup$
– pre-kidney
Apr 25 '17 at 16:33












$begingroup$
Yeah it's defininitely $x+y^2$.
$endgroup$
– Anon
Apr 25 '17 at 16:37




$begingroup$
Yeah it's defininitely $x+y^2$.
$endgroup$
– Anon
Apr 25 '17 at 16:37












$begingroup$
Okay, can you also verify that the region of integration is correct?
$endgroup$
– pre-kidney
Apr 25 '17 at 16:37




$begingroup$
Okay, can you also verify that the region of integration is correct?
$endgroup$
– pre-kidney
Apr 25 '17 at 16:37












$begingroup$
(The reason I am asking these questions is because the integral you posted appears to have no closed form expression.)
$endgroup$
– pre-kidney
Apr 25 '17 at 16:40




$begingroup$
(The reason I am asking these questions is because the integral you posted appears to have no closed form expression.)
$endgroup$
– pre-kidney
Apr 25 '17 at 16:40












$begingroup$
I've tried with this coordinate change, but it doesn't seems easy either $begin{cases} x=t^2-s^2\ y=s end{cases}$
$endgroup$
– Rafa Budría
Apr 25 '17 at 18:14






$begingroup$
I've tried with this coordinate change, but it doesn't seems easy either $begin{cases} x=t^2-s^2\ y=s end{cases}$
$endgroup$
– Rafa Budría
Apr 25 '17 at 18:14












1 Answer
1






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$begingroup$

With the substitution
begin{align*}
x &= (rcostheta)^2 \
y &= rsintheta
end{align*}
we find
begin{align*}
iint_{Omega } e^{(x+y^2)^{3/2}} ,d A
&= int_0^{pi/2}int_0^{R(theta)} 2 e^{r^3}r^2costheta ,dr,dtheta \
&= frac{2}{3} int_0^{pi/2}(e^{R(theta)^3}-1)costheta,dtheta,
end{align*}
where
$$R(theta) = frac{-sintheta+sqrt{8-7sin^2theta}}{2cos^2theta}.$$
($R(theta)$ is found by solving $x(r,theta)+y(r,theta) = 2$ for $r$.)
Letting $sintheta = t$ we find
begin{align*}
iint_{Omega } e^{(x+y^2)^{3/2}} ,mathrm{d}A
&= frac{2}{3} int_0^1 left(expleft(frac{-t+sqrt{8-7t^2}}{2(1-t^2)}right)^3 -1right)dt.
end{align*}
The final one-dimensional integral may be found numerically.






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    0












    $begingroup$

    With the substitution
    begin{align*}
    x &= (rcostheta)^2 \
    y &= rsintheta
    end{align*}
    we find
    begin{align*}
    iint_{Omega } e^{(x+y^2)^{3/2}} ,d A
    &= int_0^{pi/2}int_0^{R(theta)} 2 e^{r^3}r^2costheta ,dr,dtheta \
    &= frac{2}{3} int_0^{pi/2}(e^{R(theta)^3}-1)costheta,dtheta,
    end{align*}
    where
    $$R(theta) = frac{-sintheta+sqrt{8-7sin^2theta}}{2cos^2theta}.$$
    ($R(theta)$ is found by solving $x(r,theta)+y(r,theta) = 2$ for $r$.)
    Letting $sintheta = t$ we find
    begin{align*}
    iint_{Omega } e^{(x+y^2)^{3/2}} ,mathrm{d}A
    &= frac{2}{3} int_0^1 left(expleft(frac{-t+sqrt{8-7t^2}}{2(1-t^2)}right)^3 -1right)dt.
    end{align*}
    The final one-dimensional integral may be found numerically.






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      With the substitution
      begin{align*}
      x &= (rcostheta)^2 \
      y &= rsintheta
      end{align*}
      we find
      begin{align*}
      iint_{Omega } e^{(x+y^2)^{3/2}} ,d A
      &= int_0^{pi/2}int_0^{R(theta)} 2 e^{r^3}r^2costheta ,dr,dtheta \
      &= frac{2}{3} int_0^{pi/2}(e^{R(theta)^3}-1)costheta,dtheta,
      end{align*}
      where
      $$R(theta) = frac{-sintheta+sqrt{8-7sin^2theta}}{2cos^2theta}.$$
      ($R(theta)$ is found by solving $x(r,theta)+y(r,theta) = 2$ for $r$.)
      Letting $sintheta = t$ we find
      begin{align*}
      iint_{Omega } e^{(x+y^2)^{3/2}} ,mathrm{d}A
      &= frac{2}{3} int_0^1 left(expleft(frac{-t+sqrt{8-7t^2}}{2(1-t^2)}right)^3 -1right)dt.
      end{align*}
      The final one-dimensional integral may be found numerically.






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        With the substitution
        begin{align*}
        x &= (rcostheta)^2 \
        y &= rsintheta
        end{align*}
        we find
        begin{align*}
        iint_{Omega } e^{(x+y^2)^{3/2}} ,d A
        &= int_0^{pi/2}int_0^{R(theta)} 2 e^{r^3}r^2costheta ,dr,dtheta \
        &= frac{2}{3} int_0^{pi/2}(e^{R(theta)^3}-1)costheta,dtheta,
        end{align*}
        where
        $$R(theta) = frac{-sintheta+sqrt{8-7sin^2theta}}{2cos^2theta}.$$
        ($R(theta)$ is found by solving $x(r,theta)+y(r,theta) = 2$ for $r$.)
        Letting $sintheta = t$ we find
        begin{align*}
        iint_{Omega } e^{(x+y^2)^{3/2}} ,mathrm{d}A
        &= frac{2}{3} int_0^1 left(expleft(frac{-t+sqrt{8-7t^2}}{2(1-t^2)}right)^3 -1right)dt.
        end{align*}
        The final one-dimensional integral may be found numerically.






        share|cite|improve this answer









        $endgroup$



        With the substitution
        begin{align*}
        x &= (rcostheta)^2 \
        y &= rsintheta
        end{align*}
        we find
        begin{align*}
        iint_{Omega } e^{(x+y^2)^{3/2}} ,d A
        &= int_0^{pi/2}int_0^{R(theta)} 2 e^{r^3}r^2costheta ,dr,dtheta \
        &= frac{2}{3} int_0^{pi/2}(e^{R(theta)^3}-1)costheta,dtheta,
        end{align*}
        where
        $$R(theta) = frac{-sintheta+sqrt{8-7sin^2theta}}{2cos^2theta}.$$
        ($R(theta)$ is found by solving $x(r,theta)+y(r,theta) = 2$ for $r$.)
        Letting $sintheta = t$ we find
        begin{align*}
        iint_{Omega } e^{(x+y^2)^{3/2}} ,mathrm{d}A
        &= frac{2}{3} int_0^1 left(expleft(frac{-t+sqrt{8-7t^2}}{2(1-t^2)}right)^3 -1right)dt.
        end{align*}
        The final one-dimensional integral may be found numerically.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 6 at 1:21









        user26872user26872

        14.9k22773




        14.9k22773






























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