91Ó°ÊÓ

We need to recognize the function inside the integral as the equation for a circle in standard form, before proceeding to use a polar coordinate system. The equation given is: $$24-2 x-x^{2}=y^2$$ We need to re-arrange this equation as: $$(x-h)^2 +(y-k)^2 = r^2$$ We can rewrite the given equation as: $$x^2+2x+y^2 = 24$$ Now, complete the square for the x-terms: $$(x+1)^2-1+y^2=24$$ And rewrite the equation as: $$(x+1)^2+y^2=25$$ This equation represents a circle with center \((-1,0)\) and radius 5.

Now that we've determined that the equation represents a circle, we will convert the integral into polar coordinates. First, we need to find the relationship between cartesian and polar coordinates: $$x=r\cos(\theta), \quad y=r\sin(\theta)$$ We also need to calculate the Jacobian of the transformation: $$J = \frac{\partial(x,y)}{\partial(r,\theta)} = r$$ Now we will convert the boundaries. The x-boundaries are given as -6 and 4, so in polar coordinates, they become: $$\theta_1 = \arccos(\frac{-6+1}{5}) = \arccos(-1) = \pi$$ $$\theta_2 = \arccos(\frac{4+1}{5}) = \arccos(1) = 0$$ Now, we find the polar equation of the circle using the cartesian equation, and the polar-coordinate relations mentioned above. Substituting \(x\) and \(y\) we get, $$(r\cos(\theta)+1)^2 + (r\sin(\theta))^2 = 25$$ Solving for \(r\), we get: $$r^2\cos^2(\theta) + 2r\cos(\theta) + r^2\sin^2(\theta) = 25$$ $$r^2(\sin^2(\theta)+\cos^2(\theta)) + 2r\cos(\theta) = 25$$ $$r^2 + 2r\cos(\theta) = 25$$ Now we can set the limits for \(r\): $$0 \le r \le \sqrt{24-2 x-x^{2}}$$ So the integral becomes, $$\int\int_{A}r\,dA = \int_{\pi}^{0}\int_{0}^{\sqrt{24-2 x-x^{2}}}r\,dr\,d\theta$$

Now we can go ahead and evaluate the integral: $$\int_{\pi}^{0}\int_{0}^{\sqrt{24-2 x-x^{2}}}r\,dr\,d\theta$$ Evalute the inner integral first: $$\int_{0}^{\sqrt{24-2 x-x^{2}}}r\,dr = \frac{1}{2}r^2\Bigg|_0^{\sqrt{24-2 x-x^{2}}} = \frac{(24-2 x-x^{2})}{2}$$ Now substitute the integral back: $$\int_{\pi}^{0}\frac{(24-2 x-x^{2})}{2}\,d\theta$$ And convert it back to polar coordinates, $$\int_{\pi}^{0}\frac{(24-2 r\cos(\theta)-r^2\cos^2(\theta))}{2}\,d\theta$$ Now, evaluate the outer integral: $$\int_{\pi}^{0}\frac{(24-2 r\cos(\theta)-r^2\cos^2(\theta))}{2}\,d\theta= -\frac{r^3}{3}\cos^3(\theta)\Bigg|_{\pi}^0$$ Finally, we get: $$\int_{-6}^{4} \sqrt{24-2 x-x^{2}} d x = -\frac{r^3}{3}\cos^3(\theta)\Bigg|_{\pi}^0 = 75$$