Integrating a 2-D Function: x^2+y^2 \leq 4

[the_godfather]

Homework Statement

$$\int\int$$ y $$\sqrt{x^2+y^2}$$dx dy

Homework Equations

x$$\geq$$ 0, y$$\geq$$ 0, x^2+y^2 $$\leq$$ 4

The Attempt at a Solution

first of all, what are the limits of integration

rearranging x^2+y^2 $$\leq$$ 4 you get x = 2 - y
this would be my limit of integration for the inner integral yes?

the limits for the outer integral cannot be a function so this would go between 2 and 0?

when the first integration is performed can the square root and y be multiplied out to give

yx + y^2 or is it a case of integrating by parts/substitution?

 

[Nabeshin]

Ok first off, I think you should understand what the region is a little better. So,
$$\left[x^2+y^2\leq4\right] \neq \left[x=2-y\right]$$ that's just a simple algebra error (I assume when you took the square root you just applied it individually to 4 and y, but you can't do that!).

So, then, what does that region describe? Start off by thinking about the similar bound,
$$x^2+y^2=4$$
any idea what that is? And if you do know what it is, what kind of coordinates are very convenient for describing that kind of geometry?

 

[the_godfather]

but if you use polar co-ordinates you don't have a value for theta

r = x^2 + y^2
=4
also how can the y part be obtained?

theta = arc tan (y/x)
this bit confused me

edit: theta can be obtained from what we know that x and y > 0. therefore theta = pi/2

 

[Mark44]

Your region is not described by r = 2; it is described by 0 <= r <= 2. And theta certainly does play a role.

 

[Nabeshin]

Probably a semantics issue, but you shouldn't say theta equals anything. Theta has a range, just like Mark44 notes that r has a range, and is not simply one value.

Although, r goes from 0 to 2, not 0 to 4.

 

[the_godfather]

yes it is a semantic issue.
what i have written on paper is different to my typing. just a bad habit to always press equals on a computer.
what is my next step?

i think once i get the limits sorted i will be fine

the limits are 0<r<2
0<=theta<=pi/2

can the y factor be expressed as r sin theta?

 

[Mark44]

yes it is a semantic issue.
what i have written on paper is different to my typing. just a bad habit to always press equals on a computer.
what is my next step?

i think once i get the limits sorted i will be fine

the limits are 0<r<2
0<=theta<=pi/2

This is only 1/4 of the way around.

can the y factor be expressed as r sin theta?

Yes, but what about the rest of the integrand, and what does dx dy become?

 

[the_godfather]

x and y are both > or = 0. which to me implies only a quarter way around

dx and dy will become dr and d(theta)

$$\int$$ ^theta ₀ $$\int$$² ₀ 2r sin theta r dr d(theta)

 

[Mark44]

You're right that theta ranges between 0 and pi/2. You put your inequalities x>=0 and y>=0 in the relevant equations rather than in the problem statement, and I missed them.

Your integrand should not have 2r; the original integrand of y*sqrt(x^2 + y^2) becomes r*rsin(theta) = r^2*sin(theta). There is also the factor of r from r dr d(theta), which you correctly show.

Your iterated integral should look like this:
$$\int_{\theta = 0}^{\pi/2}~\int_{r = 0}^2 r^3~sin(\theta)~dr~d\theta$$

 

[the_godfather]

so i should get an answer of 4?

 

[Mark44]

so i should get an answer of 4?

Yep.