To compute the given integral $\iint_{\sigma} z y^2 \, dS$, we split the surface $\sigma$ into two parts, $\sigma_1$ (the cone) and $\sigma_2$ (the circular lid at $z = 4$), and calculate their contributions separately using polar coordinates. Here's the process:

Step 1: Surface Parameterization and Projections

1. Cone ($\sigma_1$):

The equation of the cone is: $z = \sqrt{x^2 + y^2}$ Using polar coordinates, we substitute: $x = r\cos\theta, \quad y = r\sin\theta, \quad z = r$ The projection onto the $xy$-plane is a circle of radius 4 since $z = 4$ at the top: [ r \in [0, 4], \quad \theta \in [0, 2\pi] ]

The normal vector magnitude for $\sigma_1$ involves the gradient: $\vec{n} = \nabla \Phi = \left(-x, -y, \sqrt{x^2 + y^2}\right), \quad |\vec{n}| = \sqrt{1 + \left(\frac{\partial z}{\partial x}\right)^2 + \left(\frac{\partial z}{\partial y}\right)^2} = \sqrt{2}.$

So, $dS_1 = \sqrt{2} \, r \, dr \, d\theta$.

The integrand becomes: $z y^2 = r (r^2 \sin^2\theta) = r^3 \sin^2\theta.$

The integral over $\sigma_1$ is: $\int_{\sigma_1} z y^2 \, dS = \int_0^{2\pi} \int_0^4 r^3 \sin^2\theta \sqrt{2} \, r \, dr \, d\theta.$

2. Circular Lid ($\sigma_2$):

At $z = 4$, the circular lid is flat, so: $z = 4, \quad x = r\cos\theta, \quad y = r\sin\theta.$

The surface element is: $dS_2 = r \, dr \, d\theta.$

The integrand becomes: $z y^2 = 4 (r^2 \sin^2\theta) = 4r^2 \sin^2\theta.$

The integral over $\sigma_2$ is: $\int_{\sigma_2} z y^2 \, dS = \int_0^{2\pi} \int_0^4 4r^2 \sin^2\theta \, r \, dr \, d\theta.$

Step 2: Evaluate Each Integral

1. Evaluate $\sigma_1$:

$\int_{\sigma_1} z y^2 \, dS = \sqrt{2} \int_0^{2\pi} \int_0^4 r^4 \sin^2\theta \, dr \, d\theta.$

Split the integral: $\int_0^{2\pi} \sin^2\theta \, d\theta = \int_0^{2\pi} \frac{1 - \cos(2\theta)}{2} \, d\theta = \pi.$

$\int_0^4 r^4 \, dr = \left[\frac{r^5}{5}\right]_0^4 = \frac{1024}{5}.$

So: $\int_{\sigma_1} z y^2 \, dS = \sqrt{2} \cdot \pi \cdot \frac{1024}{5} = \frac{1024\pi\sqrt{2}}{5}.$

2. Evaluate $\sigma_2$:

$\int_{\sigma_2} z y^2 \, dS = 4 \int_0^{2\pi} \int_0^4 r^3 \sin^2\theta \, dr \, d\theta.$

As before: $\int_0^{2\pi} \sin^2\theta \, d\theta = \pi,$ $\int_0^4 r^3 \, dr = \left[\frac{r^4}{4}\right]_0^4 = 64.$

So: $\int_{\sigma_2} z y^2 \, dS = 4 \cdot \pi \cdot 64 = 256\pi.$

Step 3: Combine Results

$\int_{\sigma} z y^2 \, dS = \int_{\sigma_1} z y^2 \, dS + \int_{\sigma_2} z y^2 \, dS = \frac{1024\pi\sqrt{2}}{5} + 256\pi.$

Would you like further simplifications or clarifications on any step?

5 Relative Questions:

1. How does the gradient of the cone contribute to the surface element calculation?
2. Why is $\sin^2\theta$ replaced using trigonometric identities during integration?
3. Can this integral be evaluated in Cartesian coordinates, and how would it differ?
4. What are some practical applications of these surface integrals?
5. How does the symmetry of the problem simplify the integration?

Tip:

Using symmetry in such problems often reduces the computational effort, especially for angular terms like $\sin^2\theta$ or $\cos^2\theta$.