# Use the divergent theorem to verify the volume of a circular cone

Let $$T$$ be a region with boundary surface $$S$$ such that $$T$$ has volume $$V=\frac{1}{3} \iint_S (x dydz +ydzdx+zdxdy)$$ use this equation to verify the volume of a circular cone with height $$h$$ and radius of base $$a$$ is $$V = \pi a^2 \frac{h}{3}$$

Here is what I got so far

Since

$$V=\frac{1}{3} \iint_S (x dydz +ydzdx+zdxdy)$$

$$V= \iint_S x dydz =\iint_S y dxdz=\iint_S z dxdy$$

Since this is a cone so I can have $$r(w,\theta)=[w\cos(\theta),w\sin(\theta),w]$$ with $$0\leq w\leq a$$ and $$0 \leq \theta \leq 2\pi$$ as the parametrization.

Now I'm stuck. I wonder if any one can give me a hint. Please.

• Div grad curl and all that, problem in Chap-1, II-26 , b.ii Commented May 11, 2021 at 11:38

Thinking geometrically can save a whole bunch of time here. Split the surface integral over the cone into two integrals, an integral over the circular base (call this surface $B$) and integral over the curved part of the cone's surface connecting base to apex (call this surface $C$).

Consider the surface integral over $C$ first. Note that the vector field whose surface integral we're evaluating is simply a scalar multiple of the radius vector. But since we placed the cone's apex at the origin, any radius vector from the origin to a point on the surface $C$ will also be a tangent vector to surface $C$ at that point. It follows that this surface integral vanishes, since the dot product of a surface tangent vector with the surface normal vector is zero at each point:

$$\iint_{C}\vec{F}\cdot\hat{n}\,\mathbb{d}S=\frac13\iint_{C}\vec{r}\cdot\hat{n}\,\mathbb{d}S=0.$$

Now for the other part of the surface integral, $\iint_{B}\vec{F}\cdot\hat{n}\,\mathbb{d}S$. Again, $\vec{F}=\frac13\vec{r}$, and since we've oriented the cone so that its axis of symmetry coincides with the $z$-axis we can immediately see that the unit normal vector to this part of the cone's surface is simply the Cartesian unit vector $$\hat{n}=\hat{z}=\langle0,0,1\rangle$$. Parametrizing this surface is also straightforward:

$$\vec{r}(\rho,\phi)=\langle\rho\cos{\phi},\rho\sin{\phi},h\rangle,~~~\text{where }0\leq\rho \leq a,0\leq\phi\leq 2\pi.$$

Hence,

$$\vec{F}\cdot\hat{n}=\frac13 h\\ \implies \iint_{B}\vec{F}\cdot\hat{n}\,\mathbb{d}S = \frac{h}{3}\iint_{B}\,\mathbb{d}S = \frac{h}{3}(\pi a^2).~~~QED$$