Using Double Integral Find the volume of sphere $x^2 + y^2 + z^2= 4 $ cut by cylinder $\ x^2+y^2=2y $

Question

Using Double Integral Find the volume of sphere $x^2 + y^2 + z^2= 4 $ cut by cylinder $\ x^2+y^2=2y $ , When i try to make integral the limits are:

$\ -1<= x<=1 $ and $\ 0<=y<=2 $ ,but i dont know what funtion i can use to find the volume.

$$\int_{-1}^1\int_0^{2} f(x,y) \,dy\, d x$$

Answer 1

When you want to find a volume of this kind of solids, just take the function to be equal to 1. Then you simply get $4$ as an answer.

Answer 2

The volume $V$ of a body $K$ is by definition $$ V=\iiint_K 1\,dx\,dy\,dz. $$ If you first integrate with respect to $z$, then, since the upper and lower limits for $z$ are the surfaces $z=\sqrt{4-x^2-y^2}$ and $z=-\sqrt{4-x^2-y^2}$ (see the picture below) you get $$ V=\iint_D\bigl(\sqrt{4-x^2-y^2}-(-\sqrt{4-x^2-y^2})\bigr)\,dx\,dy. $$

Now, what is the domain $D$? It is not given by the inequalities $-1\leq x\leq 1$ and $0\leq y\leq 2$ as you seem to imply. Instead, what one should do is that one should project the body $K$ onto the $xy$-plane orthogonally. If we look at the picture below, we find out that $D$ is exactly the disk $x^2+(y-1)^2\leq 1$.

Thus, the volume $V$ you are looking for is given by the double integral $$V=\iint_D 2\sqrt{4-x^2-y^2}\,dx\,dy$$ where $$D=\{(x,y)\in\mathbf R^2~|~x^2+(y-1)^2\leq 1\}.$$

I leave it to you to continue the calculations.

Answer 3

Indeed $f(x,y,z)=1$ $$\left\{ \begin{align} & x=r\cos \theta \\ & y=r\sin \theta \\ \end{align} \right.\,\,\,\,\,\Rightarrow \,\,\,J=r$$ we have $$-\sqrt{4-{{r}^{2}}}\le z\le \sqrt{4-{{r}^{2}}}$$ and $$\left\{ \begin{align} & {{x}^{2}}+{{y}^{2}}=4 \\ & {{x}^{2}}+{{y}^{2}}=2y \\ \end{align} \right.\,\,\,\,\,\,\,\,\,\,\Rightarrow \,\,\,\,\,\,2\sin \theta \le r\le 2$$ therefore $$V=\int_{0}^{2\pi }{\int_{2\sin \theta }^{2}{\int_{-\sqrt{4-{{r}^{2}}}}^{\sqrt{4-{{r}^{2}}}}{r\,dzdrd\theta }}}$$

Answer 4

Volume will be equal to $$V=\int_{z=-2}^2\int_{y=0}^{\frac {4-z^2}{2}}\int_{x=-\sqrt{4-y^2-z^2}}^{x=\sqrt{4-y^2-z^2}}dxdydz=\frac{112}{9}$$ hope this will be helpful.