Use Gauss' Law for Electricity to find the total charge?

Question

Use Gauss' Law for Electricity to find the total charge in the cone y=sqrt(x^2+z^2) for E=<2xy, y^2, 5xy>. (Answer: 0)

I've set up the triple integral of 4y dV and the formula for volume of a cone is V=1/3*pi*r^2*h but I can't find the radius and height.

Answer 1

Great work getting ∇·E = 4y. The issue is that the cone y = √(x² + z²) isn't a closed surface, so the divergence theorem (volume integral) approach hits a wall — you'd need a cap, and none is given. Instead, compute the flux directly through the cone surface. The beautiful result is that it vanishes by symmetry, regardless of any bounds on r.

Step 1: Verify ∇·E

$$\nabla \cdot \mathbf{E} = \frac{\partial}{\partial x}(2xy) + \frac{\partial}{\partial y}(y^2) + \frac{\partial}{\partial z}(5xy) = 2y + 2y + 0 = 4y \quad \checkmark$$

Step 2: Parameterize the Cone

The cone y = √(x² + z²) is naturally parameterized in cylindrical coordinates (using y as the cone axis):

$$\mathbf{r}(r, \theta) = (r\cos\theta,; r,; r\sin\theta), \quad r \geq 0,; 0 \leq \theta \leq 2\pi$$

(Here y = r on the surface of the cone.)

Step 3: Compute the Normal Vector

$$\mathbf{r}r = (\cos\theta,; 1,; \sin\theta), \qquad \mathbf{r}\theta = (-r\sin\theta,; 0,; r\cos\theta)$$

$$\mathbf{r}r \times \mathbf{r}\theta = (r\cos\theta,; -r,; r\sin\theta)$$

Step 4: Evaluate E on the Cone Surface

On the surface, x = rcosθ, y = r, z = rsinθ, so:

$$\mathbf{E} = \langle 2r^2\cos\theta,; r^2,; 5r^2\cos\theta \rangle$$

Step 5: Compute the Dot Product

$$\mathbf{E} \cdot (\mathbf{r}r \times \mathbf{r}\theta) = (2r^2\cos\theta)(r\cos\theta) + (r^2)(-r) + (5r^2\cos\theta)(r\sin\theta)$$

$$= 2r^3\cos^2\theta - r^3 + 5r^3\cos\theta\sin\theta$$

Step 6: Integrate Over θ (This Is the Key!)

$$\int_0^{2\pi} \left(2r^3\cos^2\theta - r^3 + 5r^3\sin\theta\cos\theta\right) d\theta$$

Using standard results:

$$\int_0^{2\pi}\cos^2\theta, d\theta = \pi, \qquad \int_0^{2\pi} d\theta = 2\pi, \qquad \int_0^{2\pi}\sin\theta\cos\theta, d\theta = 0$$

$$= 2r^3(\pi) - r^3(2\pi) + 5r^3(0) = 2\pi r^3 - 2\pi r^3 = \boxed{0}$$

The flux is zero regardless of the upper limit on r, so you never need to find a specific radius or height. The cancellation happens entirely in the θ-integration — the first and second terms perfectly cancel, and the cross-term vanishes by periodicity. That's why the answer is Q = 0.