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I want to integrate $f(x,y) = (x+y)^2 $ ont he domain $D = \{ (x,y) \in \mathbb{R^2} , x^2+y^2 ≤ 1 \}$

Now If I change to polar coordinates, I have that I want to integrate $g(r, \theta) = r^2(cos\theta + sin\theta)^2$ on the domain $D' = \{ (r,\theta) \in \mathbb{R^2} , -1 ≤ ≤ -1 , 0≤ \theta ≤ 2\pi \}$

Now if we integrate that we get :

$\int_{D'} g(r, \theta)rdrd\theta = \int_0 ^2\pi \big( \int_0 ^1 g(r, \theta)rdr\big) d\theta$

Now I don't understand why the upper and lower integrztion limits of the integral inside are $0$ and $1$ and not $-1$ and $-1$

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1 Answer 1

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If you define a unit circle with the bounds: $$0≤\theta≤2\pi,0≤r≤1$$ Then every point in the circle has a unique $\theta$ and $r$.

If you define a unit circle with the bounds: $$0≤\theta≤2\pi,-1≤r≤1$$ Then every point can be defined with two sets of $(\theta,r)$, which means the integration counts each point twice.

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  • $\begingroup$ Yes, Of course, Thank you $\endgroup$ Jan 13, 2017 at 16:24

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