how do i calcute this set into polar coordinates for my integrals? Let

$$ D_a,_b = {(x, y) \in \mathbb R^2:x \leq0, y \geq0, a^2 \leq x^2+y^2 \leq b^2} $$


Function to calculate with: $$ f(x) = 3 \sqrt{x^2+y^2} $$ so i started with: $$ \int_{\pi/2}^{\pi} \int_a^b 3 \sqrt{r^2cos(θ)^2+r^2sin(θ)^2} dr dθ$$ my current solution is: $$ -\frac{3}{4}\pi(a^2-b^2) $$

  • $\begingroup$ This is good progress. Please see update2 to my answer below. $\endgroup$ – gt6989b Jan 14 at 18:06
  • $\begingroup$ Please see another update, you did not incorporate my second remark. $\endgroup$ – gt6989b Jan 14 at 19:33

Here is a geometric approach. Your region is an annulus (area between two circles) with radii $a<b$, limited to the 2nd quadrant. Hence, $a \le r \le b$, but can you compute the range for $\theta$?

An algebraic way to see this is to note that $r^2=x^2+y^2$, so the last constraint reads $a^2 \le r^2 \le b^2$, and since $r>0$ we must have $a \le r \le b$.

Now use $x \le 0, y \ge 0$ to fix the range for $\theta$. Can you finish?


After settling our discussion in the comments, you end up with the following double integral: $$ \iint_{D_{a,b}} f(x,y) \ dx \ dy = \int_{\pi/2}^\pi \int_a^b f(r\cos \theta, r \sin \theta)\ r \ dr \ d\theta $$


Your approach is good but:

  • you can simplify the expression under the root, note that $\sin^2 \theta + \cos^2 \theta = 1$ and $\sqrt{r^2} = r$ since $r>0$
  • when you convert $dx \ dy$ to polar, you get $r \ dr \ d\theta$ (not $dr \ d\theta$ as you seem to)


Now you have $f(x,y) = 3\sqrt{x^2+y^2} = 3r$ and so you get $$ \iint_{D_{a,b}} f(x,y) \ dx \ dy = \int_{\pi/2}^\pi \int_a^b (3r) \cdot r \ dr \ d\theta = \frac{\pi}{2} \left[ b^3 - a^3 \right] $$

  • $\begingroup$ im not sure but i think i will have $$ \int_0^{2pi} \int_a^b $$ ? $\endgroup$ – Eiden Jan 14 at 14:20
  • $\begingroup$ @Eiden not quite. You are taking all possible values for $\theta$ which will use all four quadrants. You only need the second one -- where $x \le 0$ and $y \ge 0$. What range of angles does that correspond to? $\endgroup$ – gt6989b Jan 14 at 14:36
  • $\begingroup$ oh that has a range of 90 ° and so i only need pi/2 ? $\endgroup$ – Eiden Jan 14 at 15:14
  • $\begingroup$ @Eiden closer now. That does have range of $90^\circ$, but if you go $0 \le \theta \le \pi/2$ you describe the first quadrant. Can you describe the second? $\endgroup$ – gt6989b Jan 14 at 15:50
  • $\begingroup$ i guess π/2≤θ≤π ? and the other 2 are the same only for negative values ? $\endgroup$ – Eiden Jan 14 at 15:56

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