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We have two spheres whose radiuses are $a$ cm each and their centre coordinates are $(0,a,0)$ and $(0,0,0)$ consecutively.

How can we determine the volume of the intersection regions of these given two spheres by using triple integrals?

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    $\begingroup$ Adapting the ideas in the answers to Volume Between Spheres – Spherical Coordinates should answer your question. (Single-variable calculus suffices; insistence on spherical or cylindrical coordinates would be almost perverse, but an answer using spherical coordinates is given at the linked question.) $\endgroup$ – Andrew D. Hwang Jan 10 '17 at 13:54
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First note that the two spheres have equations: $$ x^2+y^2+z^2=a^2 \quad \mbox{and}\quad x^2+(y-a)^2+z^2=a^2 $$ and intersect on the plane $y=a/2$.

So the volume of the intersection region is made of two identical cups that have height $h=a/2$ and radius of the basis circle $b=\frac{\sqrt{3}}{2}a$ and we can use the formula for the volume of a cup to evaluate the volume as: $$ V=2 V_{cup}=2\cdot\frac{\pi a}{12}\left(\frac{9}{4}a^2+\frac{a^2}{4} \right)=2\cdot\frac{5\pi a^3}{24} $$

If you want to justify this result with an integral, you can do it with a simple integral noting that the single cup is a solid of revolution around the $y$ axis of the curve $z=\sqrt{a^2-y^2}$ in the $y-z$ plane. So the volume is: $$ V_{cup}=\int_{a/2}^a \pi z(y)^2 dy=\pi \int_{a/2}^a (a^2-y^2)dy= \frac{5 \pi}{24}a^3 $$

If you want necessarily a triple integral than the better is to note the cylindrical symmetry around the y axis and use cylindrical coordinates around it ( this means that we switch the name of $y$ and $z$ axis), so the volume of a single cup becomes: $$ V_{cup}=\int_{a/2}^a\int_0^{2\pi}\int_0^{\sqrt{a^2-y^2}}\rho \;d \rho\; d \varphi\; d y $$

that, with a bit of work, gives the same result.

The use of a triple integral in rectangular coordinates is possible but really masochistic.

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