Double integral - finding the region I need to find the volume of an object bounded by the following planes:
$$z=y+4$$
$$x^{2}+y^{2}=4$$
$$z=0$$
Is this true that the region of interest on the XY plane is the whole circle drew by the second formula above?
 A: If you do a little sketch, you get
$$-2\le x\le 2\;,\;\;-\sqrt{4-x^2}\le y\le\sqrt{4-x^2}\;,\;\;0\le z\le y+4\implies$$
$$\int\limits_{-2}^2dx\int\limits_{-\sqrt{4-x^2}}^{\sqrt{4-x^2}}dy\int\limits_0^{y+4}dz=\int\limits_{-2}^2dx\int\limits_{-\sqrt{4-x^2}}^{\sqrt{4-x^2}}(y+4)dy=$$
$$=\int\limits_{-2}^2 8\sqrt{4-x^2}=16\int\limits_{-2}^2\sqrt{1-\left(\frac x2\right)^2}\;dx=32\int\limits_{-\pi/2}^{\pi/2}\cos^2t\,dt$$
The last equality following from the substitution
$$\frac x2=\sin t\;,\;\;dx=2\cos t\,dt\;,\;\;\text{ so }$$
$$32\int\limits_{-\pi/2}^{\pi/2}\cos^2t\,dt=\left.32\frac{t+\sin t\cos t}2\right|_{-\pi/2}^{\pi/2}=16\pi$$
A: Note that $x^2+y^2=4$ is an infinite cylinder. Note that $z=0$ is the $xy$-plane, and that $z=y+4$ is a plane with slope $1$ and lying paralell to the $x$ four units above.
Try to integrate it through the $zy$-plane, that is, integrate the areas of all the trapezoids that arise when you cross cut it. Alternatively, you can integrate rectangles using the $xz$-plane.
Using this last approach, I am getting the volume should be $$2\int_{-2}^2 (4+\alpha)\sqrt{4-\alpha^2}d\alpha$$
Since the rectangle will have height $4+\alpha$ and width $2\sqrt{4-\alpha^2}$ when we cut it with a plane paralell to the $xz$-plane at $-2\leq \alpha\leq 2$. The substitution $\alpha=2\sin\theta$ seems like the most natural next step.
ADD Note that because one part of the integrand is even, we get $$\begin{align}
  2\int_{ - 2}^2 {(4 + \alpha )} \sqrt {4 - {\alpha ^2}} d\alpha  &= \underbrace {2\int_{ - 2}^2 {\alpha \sqrt {4 - {\alpha ^2}} d\alpha } }_{ = 0} + 8\int_{ - 2}^2 {\sqrt {4 - {\alpha ^2}} d\alpha } \\ 
   &= 8\int_{ - 2}^2 {\sqrt {4 - {\alpha ^2}} d\alpha } \\ 
   &= 8\int_{ - 2}^2 {\sqrt {4 - {\alpha ^2}} d\alpha }  = 8\frac{{4\pi }}{2}\\ 
   &= 16\pi  \end{align} $$
which agrees with Antonio's answer. Note we use that $$\int_{ - r}^r {\sqrt {{r^2} - {\alpha ^2}} d\alpha }  = \frac{{\pi {r^2}}}{2}$$
A: I'd first change the function to cilindrical coordinates:
$$x = r \sin(\theta), y = r \cos(\theta)$$
$$0 < r < 2$$
$$0 < \theta < 2\pi$$
Then integrate it over the x-y plane, resulting in circles, followed by r & $\theta$
$$\int_0^{2\pi}\int_0^2\int_0^{r\cdot \sin(\theta)+4}1 \cdot r \space dzdrd\theta$$
$$=\int_0^{2\pi} \int_0^2 r^2 \sin(\theta)+4r \space drd\theta$$
etc...
A: $$0=y+4$$
$$y=-4$$
Since $y=-4$ lies outside the radius $2$ circle, the plane $z=y+4$ does not bisect the circle on the xy plane. So the interval of integration (using cylindrical coordinates) is $0\le{r}<2$ and $0\le\theta<2\pi$:
$$\int_{0}^{2\pi}\int_{0}^{2}(r\sin\theta+4)r\space{dr}\space{d\theta}$$
