Can This Double Integral Be Solved in Closed Form?

[mabauti]

is there a closed form solution for this double integral?

$\int^{2}_{1}$$\int^{3}_{4}$$\sqrt{1+4x^{2}+4y^{2}}dydx$

 

[jedishrfu]

is there a closed form solution for this double integral?

$\int^{2}_{1}$$\int^{3}_{4}$$\sqrt{1+4x^{2}+4y^{2}}dydx$

Have you tried polar coordinates?

 

[HallsofIvy]

Exactly what do you mean by "closed form solution"? Since a limits of integration are numbers, the integral is, of course, a single number, which is about as "closed form" as you can get! But I expect that you are asking about a method of finding that number, which is a different matter.

As jedishrfu said, seeing that $x^2+ y^2$, I would try polar coordinates. Of course, the fact that the region of integration is a rectangle rather than a disk complicates things!

As the "$\theta$" line sweeps up from the x-axis ($\theta= 0$) it first touches the rectangle at (2, 3) (slope 3/2), then at the point (2, 4) (slope 2), then (1, 3) (slope 3), and finally at (1, 4) (slope 4). That gives three intervals, $\theta= arctan(3/2)$ to $arctan(2)$, $\theta= arctan(2)$ to $arctan(3)$, and $\theta= arctan(3)$ to $arctan(4)$ on which the ray enters and leaves the rectangle through different edges.

For $\theta= arctan(3/2)$ to $arctan(2)$, the ray enters the rectangle through the bottom, $y= r sin(\theta)= 3$ (so $r= 3/sin(\theta)= 3 csc(\theta)$) and leaves through the right side, $x= rcos(\theta)= 2$ (so $r= 2/cos(\theta)=2 sec(\theta)$. That integral will be
$$\int_{arctan(3/2)}^{arctan(2)}\int_{3csc(\theta)}^{2sec(\theta)}\sqrt{1+4r^2} rdrd\theta$$.

For $\theta= arctan(2)$ to $arctan(3)$, the ray enters the rectangle through the bottom and leaves through top, $y= r sin(\theta)= 4$ (so $r= 4/sin(\theta)= 4 csc(\theta)$). That integral will be
$$\int_{arctan(2)}^{arctan(3)}\int_{3csc(\theta)}^{4csc(\theta)}\sqrt{1+4r^2} rdrd\theta$$.

Finally, for $\theta= arctan(3)$ to $arctan(4)$, the ray enters the rectangle through the left edge, $x= r cos(\theta)= 1$ (so $r= 1/cos(\theta)= sec(\theta)$). That integral will be
$$\int_{arctan(3)}^{arctan(4)}\int_{3sec(\theta)}^{4csc(\theta)}\sqrt{1+4r^2} rdrd\theta$$.

Now, can you do those integrals? Would you consider that "closed form"?

 

[mabauti]

@ jedishrfu : I tried , but I couldn't figure out the intervals for theta/r

@ HallsofIvy: it worked, I can figure out the intervals for theta, but not for r. Thanks :)

 

[pasmith]

@ jedishrfu : I tried , but I couldn't figure out the intervals for theta/r

@ HallsofIvy: it worked, I can figure out the intervals for theta, but not for r. Thanks :)

The limits of the r integral are functions of theta, so you have to do the r integral before the theta integral. And the r integral is straightforward: substitute $u = 1 + 4r^2$.

 

[mabauti]

??

I did solve the integrals.

pasmith: could you elaborate please?