Surface Integrals for Calculating Volume

Question

I understand that volume under a surface can be calculated with double integrals in Multivariable Calculus, but can it also be calculated with surface integrals? I would think that taking the surface integral of the height of the surface would give its volume. E.g., I assumed that ∬ z dS where S is the upper half of a radius 2 sphere would give the volume of that region, but the answer was apparently 8$\pi$. Why doesn't this integral give the volume of the region?

Answer 1

I understand that volume under a curve can be calculated with double integrals

Observe that the (signed) volume $$\iint_L f(x,y) \:\mathrm dx\,\mathrm dy\tag1$$ involves the surface $z=f(x,y)$ lying in 3D space and the lamina $L$ lying in 2D space.

I assume the boldfaced text is a typo and you mean “volume under a surface” (analogous to area under a curve) instead.

I would think that taking the surface integral of the height of the surface would give its volume. E.g., I assumed that $$∬_S z \:\mathrm dS,\tag2$$ where $S$ is the upper half of a radius-$2$ sphere, would give the volume of that region, but the answer was apparently $8\pi$. Why doesn't this integral give the volume of the region?

a way to find the signed area under a surface using the surface integral. Is it just not possible?

Another typo: again, I assume you mean “volume under a surface” instead.

Here (I suppose you are referring to this Example 2), the integration domain $S$ is a surface residing in 3D space, and the integrand is a function $f(x,y,z)$ on $\mathbb R^3$ of value $z,$ which is neither $S$'s height nor gives a surface or curve.

(To be clear: integral $(1)$'s integrand gives a surface, while integral $(2)$'s integration domain is a surface. Integral $(2)$'s integrand is a 3D scalar field, while integral $(1)$'s integrand is a 2D scalar field.)

Since $\int_S \,\mathrm dS$ gives the area of $S$ which already occupies 3D space, $\int_S f(x,y,z)\,\mathrm dS$ does not in general compute a volume associated with either the concave or convex side of $S.$ Generally though, if $f(x,y,z)$ gives $S$'s density, then $\int_S f(x,y,z)\,\mathrm dS$ gives its mass.

More here: What do these integrals compute?

Answer 2

If $R$ is a regular bounded region in $\mathbb{R}^3$ and $\partial R$ is its boundary (a closed surface), the divergence theorem states that $$ \iiint_R \textrm{div} F\, dV = \iint_{\partial R} F\cdot n \,dS $$

So, you can take any vector field with divergence equal to 1, for instance $F(x,y,z)=(x,0,0)$, and compute $$ Vol(R) = \iint_{\partial R} F \cdot n \, dS $$