Let's solve the problem step by step: ### **Given:** Evaluate the surface integral: \[ \iint_S y^2 z^2 \, dS \] where \( S \) is the part of the cone \( y = \sqrt{x^2 + z^2} \) for \( 0 \leq y \leq 5 \). --- ## **Step 1: Parametrize the Surface** The cone is given by \( y = \sqrt{x^2 + z^2} \). Let’s use cylindrical coordinates: - \( x = r \cos\theta \) - \( z = r \sin\theta \) - \( y = r \) - \( 0 \leq r \leq 5 \), \( 0 \leq \theta < 2\pi \) So, the parametrization is: \[ \mathbf{r}(r, \theta) = (r \cos\theta, r, r \sin\theta) \] --- ## **Step 2: Compute \( dS \) (Surface Element)** For a surface given by \( \mathbf{r}(u,v) \), the surface element is: \[ dS = |\mathbf{r}_r \times \mathbf{r}_\theta| dr d\theta \] Compute the partial derivatives: \[ \mathbf{r}_r = (\cos\theta, 1, \sin\theta) \] \[ \mathbf{r}_\theta = (-r\sin\theta, 0, r\cos\theta) \] Cross product: \[ \mathbf{r}_r \times \mathbf{r}_\theta = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ \cos\theta & 1 & \sin\theta \\ -r\sin\theta & 0 & r\cos\theta \end{vmatrix} \] \[ = (1 \cdot r\cos\theta - \sin\theta \cdot 0)\mathbf{i} - (\cos\theta \cdot r\cos\theta - \sin\theta \cdot (-r\sin\theta))\mathbf{j} + (\cos\theta \cdot 0 - 1 \cdot (-r\sin\theta))\mathbf{k} \] \[ = (r\cos\theta)\mathbf{i} - (r\cos^2\theta + r\sin^2\theta)\mathbf{j} + (r\sin\theta)\mathbf{k} \] \[ = (r\cos\theta, -r(\cos^2\theta + \sin^2\theta), r\sin\theta) \] \[ = (r\cos\theta, -r, r\sin\theta) \] Magnitude: \[ |\mathbf{r}_r \times \mathbf{r}_\theta| = \sqrt{(r\cos\theta)^2 + (-r)^2 + (r\sin\theta)^2} \] \[ = \sqrt{r^2\cos^2\theta + r^2 + r^2\sin^2\theta} \] \[ = \sqrt{r^2(\cos^2\theta + \sin^2\theta) + r^2} \] \[ = \sqrt{r^2(1) + r^2} \] \[ = \sqrt{2r^2} \] \[ = r\sqrt{2} \] --- ## **Step 3: Write the Integrand in Terms of \( r, \theta \)** Given \( y^2 z^2 \): - \( y = r \) - \( z = r \sin\theta \) So: \[ y^2 z^2 = (r^2)(r^2 \sin^2\theta) = r^4 \sin^2\theta \] --- ## **Step 4: Set Up the Integral** \[ \iint_S y^2 z^2 \, dS = \int_{0}^{2\pi} \int_{0}^{5} r^4 \sin^2\theta \cdot r\sqrt{2} \, dr \, d\theta \] \[ = \sqrt{2} \int_{0}^{2\pi} \int_{0}^{5} r^5 \sin^2\theta \, dr \, d\theta \] --- ## **Step 5: Integrate** ### Integrate with respect to \( r \): \[ \int_{0}^{5} r^5 \, dr = \left[ \frac{r^6}{6} \right]_{0}^{5} = \frac{5^6}{6} = \frac{15625}{6} \] ### Now the \( \theta \) Integral: \[ \sqrt{2} \times \frac{15625}{6} \int_{0}^{2\pi} \sin^2\theta \, d\theta \] Recall: \[ \int_{0}^{2\pi} \sin^2\theta \, d\theta = \pi \] (because \( \sin^2\theta = \frac{1-\cos2\theta}{2} \), so integral over \( 0 \) to \( 2\pi \) gives \( \pi \)) --- ## **Step 6: Final Answer** \[ \text{Final Answer} = \sqrt{2} \times \frac{15625}{6} \times \pi = \frac{15625\sqrt{2}\pi}{6} \] --- ### **Boxed Final Answer:** \[ \boxed{ \frac{15625\sqrt{2}\pi}{6} } \]

The surface integral over the cone is evaluated by parametrizing the surface in cylindrical coordinates, calculating the surface element via the cross product of partial derivatives, and expressing the integrand in terms of these parameters. The integral reduces to standard integrals over \( r \) and \( \theta \), resulting in the final value \(\frac{15625 \sqrt{2} \pi}{6}\).