In calculating the volume bounded by cylinders $x^2 + y^2 = r^2$ and $z^2 + y^2 = r^2$, the following equation is usually derived:
$$\iint\limits_{D} f(x,y) \ dA = \int\limits_{-r}^{r}\int\limits_{-\sqrt{r^{2}-y^{2}}}^{\sqrt{r^{2}-y^{2}}} f(x,y) \ dx dy$$
leading to $$\int\limits_{-r}^{r}\int\limits_{-\sqrt{r^{2}-y^{2}}}^{\sqrt{r^{2}-y^{2}}} 2\sqrt{r^{2}-y^{2}} \ dx dy = \frac{16}{3}r^{3}$$
But I cannot understand why there is $2$ in front of $\sqrt {r^2-y^2}$. Can anyone tell me why $2$ is needed?
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$\begingroup$Consider the following handy plot. Let find the volume of the intersection as it is shown. Can you find it? I consider the part in $0\le x,y\leq a$, so you may need multiplying it by $8$. That square has the value of $a^2-x^2$ as its area(Why?). Note that $C(x,0,0),~~D(x, \sqrt{a^2-x^2},0)$, so we have the selected volume as $$V=\int_0^a(a^2-x^2)dx$$
The simple way to solve this is by coherent units. The figure described is a rototope ([x,z],y).
The area of x and z is 1 square unit. The volume of [x,y] and z under crind product is the product of their crind volumes, ie $\frac 4{\pi}$ and $1$, this gives $\frac 4{\pi}$ spherical units. This is converted into cubic units at one sphere = $\frac {\pi}6$ cubic units, gives $\frac 2 3$ of the containing cube.
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