# How To Convert A Set Into Polar Coordinates For Integration?

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

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

Thanks!

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: $$-\frac{3}{4}\pi(a^2-b^2)$$

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

Here is a geometric approach. Your region is an annulus (area between two circles) with radii $$a, limited to the 2nd quadrant. Hence, $$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?

UPDATE

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

UPDATE 2

• 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 $$\iint_{D_{a,b}} f(x,y) \ dx \ dy = \int_{\pi/2}^\pi \int_a^b (3r) \cdot r \ dr \ d\theta = \frac{\pi}{2} \left[ b^3 - a^3 \right]$$
• im not sure but i think i will have $$\int_0^{2pi} \int_a^b$$ ? – Eiden Jan 14 at 14:20
• @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? – gt6989b Jan 14 at 14:36
• @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? – gt6989b Jan 14 at 15:50