# Integrating double integral with spherical coordinates problem with interpret a domain

Hello i have the following problem i am solving integral with spherical coordinates but i am getting wrong answer - i think i am integrating correct so i think the problem is coming from the constraints. So i have $$\iint_Dxdx$$ and i have $$D = (x^2 + y^2 \le 2y, x \ge 0, y \ge \frac{1}{2})$$. So i begin solving - first i complete the circle and came with $$(x-0)^2 + (y-1)^2 = 1$$ that is circle centered at $$y = 1$$ with $$radius$$ = 1 so i substitute $$x^2 + y^2 = 2y$$ with $$r^2 = 2rsin\theta$$ i divide both sides by r i get $$r = 2sin\theta$$ so i have limits from $$0$$ to $$2sin\theta$$ for r also i need to rotate $$\phi$$ from $$0$$ to $$\pi$$ because that is how the circle is placed. And i got $$\int_0^\pi \int_0^{2sin\theta} (rcos\theta)r \,d\varphi\,dr$$ and so after integrating $$r$$ i am left with $$\frac{1}{3}\int_o^\pi sin^3\theta cos\theta$$ i am using u substitution and say $$u = sin\theta$$ and $$du = cos\theta$$ so i have $$\frac{8}{3}\int_0^\pi u^3du$$ and the new limits are from $$0$$ to $$0$$ because $$sin\theta$$ from 0 is 0 and $$sin\theta$$ from $$\pi$$ is also $$0$$ that means the final answer is zero in and it should be $$\frac {9}{16}$$. I am not good at math so i cannot see where my error is. Thank you for any help in advance.

The limits are $$x^2+y^2\le 2y \ ,\ x\ge0 \ , y\ge\frac{1}{2}$$

You've found the upper limit i.e, $$r = 2\sin\theta$$ (upper limit).

For $$y\ge\frac{1}{2} \ , \ r\sin\theta \ge\frac{1}{2}$$

So, the lower limit of $$r$$ is $$\frac{1}{2\sin\theta}$$

Also, $$x\ge0$$ , $$r\cos\theta\ge0$$ , $$\cos\theta\ge0$$ So, $$\theta\le\pi/2$$

At $$y=\frac{1}{2}$$, $$\sin\theta = \frac{\pi}{6}$$ [as, $$1.\sin(\pi/6) =1/2$$] $$I = \int^{\pi/2}_{\pi/6}\int^{2\sin\theta}_{\frac{1}{2sin\theta}}r^2\cos\theta \ dr d\theta = \frac{1}{3}\int^{\pi/2}_{\pi/6}\bigg\{8\sin^3\theta - \frac{1}{8sin^3\theta}\bigg\}\cos\theta d\theta$$

Now let $$u = sin\theta$$ , $$du = \cos\theta d\theta$$

At $$\theta = \pi/6$$, $$u=1/2$$ and at $$\theta = \pi/2$$, $$u=1$$

So, $$I = \frac{1}{3}\int^{1}_{1/2}\bigg\{8u^3- \frac{1}{8u^3}\bigg\}du = \frac{1}{3}\bigg[8\cdot\frac{1}{4}(1-\frac{1}{16}) - \frac{1}{8}\cdot\frac{-1}{2}\bigg(1 - \frac{1}{(1/2)^2}\bigg)\bigg] = \frac{1}{3}$$

$$I = \frac{1}{3}\big[ \frac{15}{8} - \frac{3}{16}\big] = \frac{1}{3}[\frac{27}{16}] = \frac{9}{16}$$

• Thank you very much for helping me . I perfectly get why the lower limit for $\phi$ is $pi/6$ we just substitute the angle for y into the domain restrictions but why when $cos\theta \ge 0$ we get $\theta < \frac{pi}{2}$ also why 1/2sin is the lower limit for theta – Boris Borovski May 23 at 6:49
• $\cos\theta$ is positive in the first quadrant (here we consider 1st and 2nd quadrants only) So $\theta\le\pi/2$ and as$y\ge1/2$, $r\sin\theta\ge1/2$ or $r\ge \frac{1}{2\sin\theta}$ – Ak19 May 23 at 6:51
• Thank you very much again i got it perfectly now. – Boris Borovski May 23 at 6:57
• You're welcome! – Ak19 May 23 at 6:58