Multiple integration

Question

We know that the double integration is used to find the area of a planer region and volume of a solid object in space. My question is why we use triple integration to find again the volume a body? (I know the importance of triple integration, any boddy plz)

Answer 1

A double integral works for finding volume of a 3D region only when you can establish that the upper and lower bounds (in the $z$ coordinate) can be defined by functions that only depend on $x$ and $y$.

Given a function $f(x,y)$ and a region $R$ in the $x,y$ plane, the volume of the solid bounded between the surface defined by $f$, the place, and $R$ is $$\iint_Rf(x,y)\,\mathrm dA$$ For simplicity, let's agree to use a rectangular region for $R$ with $x\in(a,b)$ and $y\in(c,d)$, so that the above can be written $$\int_c^d\int_a^bf(x,y)\,\mathrm dx\,\mathrm dy$$ Denote this solid by $S$. This double integral is equivalent to the triple integral $$\iiint_S\mathrm dV=\int_c^d\int_a^b \int_0^{f(x,y)} 1\,\mathrm dz\,\mathrm dx\,\mathrm dy$$ But this equivalence only holds if the bounding surface $f$ is an explicit function of the real plane.

The example from my comment challenges you to use a double integral to compute a sphere's volume. Without relying on the symmetry argument I alluded to, this is difficult. A sphere (radius $1$, centered at the origin) is defined by the implicit function $x^2+y^2+z^2=1$. There is no way to write $z$ explicitly as the subject of $x$ and $y$. You could use $z=\pm\sqrt{1-x^2-y^2}$ as the limits of integration with respect to $z$, but that only works in this case because this particular sphere can be cut perfectly in half along the $x,y$ plane Not so for more complicated regions, or even a sphere that is translated a little in either vertical direction.