Integrating Double Integrals: A Calculus 1 Challenge

[Mentallic]

Homework Statement

find \int\int_D sin\left(\frac{y}{x}\right)dA bounded by x=0, y=\pi, x=y^2

The Attempt at a Solution

I've only studied calculus 1, this problem is for my friend. I did read up briefly on double integrals however and this is why I'm stuck:

From the limits and where the graphs intersect, we have:

\int_0^{\pi^2}\int_{\sqrt{x}}^{\pi}sin\left(\frac{y}{x}\right)dydx

then integrating and evaluating the inside part:

\int_0^{\pi^2}\left(-xcos\left(\frac{\pi}{x}\right)+xcos\left(\frac{1}{\sqrt{x}}\right)\right)dx

But finding the integral of that seems impossible. I also tried reversing the order of integration, but come up with the same problem.

 

[gabbagabbahey]

then integrating and evaluating the inside part:

\int_0^{\pi^2}\left(-xcos\left(\frac{\pi}{x}\right)+xcos\left(\frac{1}{\sqrt{x}}\right)\right)dx

But finding the integral of that seems impossible. I also tried reversing the order of integration, but come up with the same problem.

I think you'll have to make use of the special function

\text{Ci}(z)\equiv-\int_z^\infty\frac{\cos(t)}{t}dt

called the Cosine integral


Just split the integral into two and make the substitution u=\frac{\pi}{x} for the first, and u=\frac{1}{\sqrt{x}} for the second.

 

[Mentallic]

I think you'll have to make use of the special function

\text{Ci}(z)\equiv-\int_z^\infty\frac{\cos(t)}{t}dt

called the Cosine integral


Just split the integral into two and make the substitution u=\frac{\pi}{x} for the first, and u=\frac{1}{\sqrt{x}} for the second.

Ok, let me try this...

u=\frac{\pi}{x}, x=\frac{\pi}{u}, dx=\frac{-\pi}{u^2}du

v=\frac{1}{\sqrt{x}}, x=\frac{1}{v^2}, dx=\frac{-2}{v^3}dv

So substituting all this in:

\pi^2\int_0^{\pi^2}\frac{cosu}{u^3}du+2\int_{\pi^2}^{0}\frac{cosv}{v^5}dv

But I'm unsure how to apply the fact that Ci(z)=\int\frac{cosz}{z}dz to this expression.

 

[gabbagabbahey]

So substituting all this in:

\pi^2\int_0^{\pi^2}\frac{cosu}{u^3}du+2\int_{\pi^2}^{0}\frac{cosv}{v^5}dv

But I'm unsure how to apply the fact that Ci(z)=\int\frac{cosz}{z}dz to this expression.

First, you need to change your limits of integration, since ytou are no longer integrating over x. Second, there's no need too use two different variables since \int_a^b f(v)dv=\int_a^b f(u)du (i.e. for definite integral, the integration variable is essentially a dummy variable)

You should end up with

\int\int_D \sin\left(\frac{y}{x}\right)dA=-\pi^2\int_{\frac{1}{\pi}}^\infty\frac{\cos u}{u^3}du+2\int_{\frac{1}{\pi}}^\infty\frac{\cos u}{u^5}du

Use integration by parts twice on each.

 

[Mentallic]

Oh yes of course, thanks that's really helpful

Obviously integrating by parts will give a big long expression and looking at it now, I'm not even going to think about posting it here. Taking \infty as one of the limits of the integration cancels out a lot.

But one more thing, can cos\left(\frac{1}{\pi}\right) be expressed more simply? And how do I evaluate the Ci\left(\frac{1}{\pi}\right)?

 

[gabbagabbahey]

But one more thing, can cos\left(\frac{1}{\pi}\right) be expressed more simply?

No.

And how do I evaluate the Ci\left(\frac{1}{\pi}\right)?

Same way you evaluate e (the base of the natural logarithm) or \pi; approximate it numerically to arbitrary precision using a power series. In other words, just leave it as is. It can't be simplified in terms of elementary functions.

 

[Mentallic]

Yes of course, but it's just that I don't know what the power series for Ci(z) is and I'm afraid my calculator doesn't have a button for it either

And sometimes including numerical approximations are a nice addition to the big long expression.

 

[gabbagabbahey]

Just use a better calculator

I'm sure Wolfram alpha (Google it) will have no problem giving you a numerical approximation. Alternatively, I'd venture a guess that you could look up the power series online and find it quickly.