Double Integral in Polar Coordinates problem

[DieCommie]

I am having trouble with this seemingly easy problem.

Evaluate the double integral (sin(x^2+y^2)) , where the region is 16=<x^2+y^2=<81.

I find the region in polar coordinates to be 4=<r=<9 0=<theta=<2pi
I find the expression to be sin(rcos^2theta+rsin^2theta) r dr dtheta , which is equal to sin(r) r dr dtheta

Is what I have done right? I evaluate the integral, first with respect to r over using 4 and 9, then with respect to theta using 0 and 2pi. The answer I get is ~42.43989 but that is wrong

any ideas, hints would be greatly appreciated thanks :tongue2:

 

[DieCommie]

Ill try using this "tex" to make it more understandable

$$\int \int _R sin(x^2+y^2) dA$$ where R is the region $$16 \leq x^2 + y^2 \leq 81$$

I transfer to polar coord. and get the region $$4 \leq r \leq 9$$ and $$0 \leq \theta \leq 2pi$$ and the integral $$\int^{2pi} _0 \int^9 _4 sin(r) r dr d\theta$$

I solve the inner integral using the parts method and get $$[-9cos(9)+sin(9)]-[-4cos(4)+sin(4)]$$ then i solve the outer integral and get $$2\pi [[-9cos(9)+sin(9)]-[-4cos(4)+sin(4)]]\approx 42.44$$

 

[nrqed]

Ill try using this "tex" to make it more understandable

$$\int \int _R sin(x^2+y^2) dA$$ where R is the region $$16 \leq x^2 + y^2 \leq 81$$

I transfer to polar coord. and get the region $$4 \leq r \leq 9$$ and $$0 \leq \theta \leq 2pi$$ and the integral $$\int^{2pi} _0 \int^9 _4 sin(r) r dr d\theta$$

I solve the inner integral using the parts method and get $$[-9cos(9)+sin(9)]-[-4cos(4)+sin(4)]$$ then i solve the outer integral and get $$2\pi [[-9cos(9)+sin(9)]-[-4cos(4)+sin(4)]]\approx 42.44$$

The argument of your sin function is r^2, not r.

So you will need to integrate sin(r^2) r dr. Just define a new variable, say, t=r^2. Then r dr = dt/2 so you will integrate $\int r dr sin(r^2) = {1 \over 2} \int sin(t) dt$...Simple, isn't?

Patrick

 

[DieCommie]

yes, very simple thank you.

Something I have been wondering about, that my teacher didnt go into much, is why do we add the extra $$r$$ in the integral after converting to polar coord.?

 

[Hurkyl]

There are several ways of looking at it.

The geometric idea is that you're doing an area integral... so what is the area of the small region bounded by $r \in [r_0, r_0 + \Delta r_0]$ and $\theta \in [\theta_0, \theta_0 + \Delta \theta_0]$?

Algebraically, it's just the chain rule, either via the Jacobian, or via arithmetic with differential forms.