I have to calculate the volume of the following region:
$$D=\{(x,y,x)\in \mathbb{R}^3: x^2+4y^2 \leq \pi^2, \quad x^2 +y^2 > \frac{\pi^2}{4} ,\quad x>0, \quad 0<z<x\cdot \cos(y)\}.$$ The projection of the region on the $xy$ plane is the region between an ellipse and a circle, so my attempt was to calculate a triple integral exploiting the fact that the domain is normal with respect to the $xy$ plane:
$$V=\iiint_D dxdydz=\int\int_A \left(\int_{z=0}^{z=x \cdot \cos(y)}dz\right )dxdy$$
where $$A=\{(x,y)\in \mathbb{R}^2: -\frac{\pi}{2}\leq y\leq \frac{\pi}{2},\sqrt{\frac{\pi^2}{4}-y^2}\quad \leq x\leq \sqrt{\pi^2-4y^2} \}$$
After a bit of algebra I arrived to the conclusion that: V=$6$. Now I don't know if the result is correct.
Best Answer
$$\int_0^{x \cos y} \, dz=x \cos y$$ $$\int_{\frac{1}{2} \sqrt{\pi ^2-4 y^2}}^{\sqrt{\pi ^2-4 y^2}} x \cos y \, dx=\frac{3}{8} \left(\pi ^2-4 y^2\right) \cos y$$ And finally $$\int_{-\frac{\pi }{2}}^{\frac{\pi }{2}} \frac{3}{8} \left(\pi ^2-4 y^2\right) \cos y \, dy=6$$
Last integral can be solved by parts $$\int \frac{3}{8} \left(\pi ^2-4 y^2\right) \cos y \, dy=\frac{3}{8} \left(\pi ^2-4 y^2\right) \sin y-\int \left(-3 y \sin y \right)\, dy=$$ $$=\frac{3}{8} \left(\pi ^2-4 y^2\right) \sin y-\left(3y \cos y-\int 3\cos y\,dy\right)=$$ $$=\frac{3}{8} \left(\pi ^2-4 y^2\right) \sin y-3y\cos y+3\sin y+C=$$ $$=\frac{3}{8} \left(8-4 y^2+\pi ^2\right) \sin y-3 y \cos y+C$$ Thus the last definite integral becomes $$\int_{-\frac{\pi }{2}}^{\frac{\pi }{2}} \frac{3}{8} \left(\pi ^2-4 y^2\right) \cos y \, dy=\left[\frac{3}{8} \left(8-4 y^2+\pi ^2\right) \sin y-3 y \cos y\right]_{-\pi/2}^{\pi/2}=3-(-3)=6$$