Question

Find the center of mass of the lamina that occupies the region D = {(x, y) : 4 ≤ x 2 + y 2 ≤ 16 and y ≥ |x|} with density function rho(x, y) = y + e √ x2+y 2 .

Answer 1

Answer:

$56*[e\frac{\pi}{2} +\sqrt{2}]/3$

Explanation:

Lamina that looks like a piece of a disk.

This is the expression as written:$\\ \rho (x,y)= y+e*\sqrt{x^{2} +y^{2} }$

Where e is considered a constant. So, for the second condition see first image. It is the upper zone.

Now, the first condition, it is the zone with red and green combination (not the only red nor only green).

So, combining the two conditions, would be the disk with the angles between π/4 and 3π/4. (third image)

Now, knowing the zone, we proceed with the integral, using polar coordinates :

$x^{2} +y^{2} =r^{2} \\ 4\leq r^{2} \leq 16$

$y=rsin(\theta )$

$\int\limits^\frac{3\pi }{4}_\frac{\pi }{4}  {\int\limits^4_2 {(rsin(\theta )+e*r)*r} \, dr } \, d \theta$

$\int\limits^\frac{3\pi }{4}_\frac{\pi }{4}  {\int\limits^4_2 {r^2(sin(\theta )+e)} \, dr } \, d \theta$

$\int\limits^\frac{3\pi }{4}_\frac{\pi }{4}  {\frac{56}{3} (sin(\theta )+e)} \, d \theta\\ =\frac{56}{3} (-cos(\theta)+e*\theta)\left \{ {{\theta=3\pi /4} \atop {\theta=\pi/4}} \right.$

$=56*[e\frac{\pi}{2} +\sqrt{2}]/3$