compute J=∫∫√(x^2+y^2)dx dy. x^2+y^2≤4. i've convert the integral to polar coordinates and i've got ∫∫ρ2dθdρ. but i'm stuck. how do i convert x^2+y^2≤4?

Question

loser66 · Answer

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any idea?

anonymous · Answer

nope

loser66 · Answer

why? you study double integral, at least, you have "something"  to solve the problem, right? Cannot solve if you don't know anything, If I work by myself, how can you understand mine with that "nope"?

zepdrix · Answer

Hmm have you tried converting to polar? :)

anonymous · Answer

no i didnt. i'll try now

zepdrix · Answer

I think it will work out really nicely.
Lemme know if you get stuck.

anonymous · Answer

i dont have the answer for the problem to compare. for the second integral i noticed from the book that it should have 0 and pi buy i don't understand how the first one needs to be. can you please explain it to me? i've found more examples and neither had a pattern that i could follow

john_es · Answer

As you have said, you have the condition,
$$x^2+y^2\leq 4$$
When you apply the polar change, then, it should be,
$$0\leq \rho^2\leq 4\Rightarrow 0\leq \rho\leq 2$$
So (now corrected)
$$\int_0^2\int_0^{2\pi}\rho^2d\rho d\theta=\frac{2\pi}{3}\rho^3=\frac{16\pi}{3}$$

anonymous · Answer

thank you so much.