Question

Let \( S \) be the part of the cone \( z^2 = x^2 + y^2 \) between the planes \( z = 0 \) and \( z = 1. \) Then the value of the surface integral \( \iint_S (x^2 + y^2) \, dS \) is

• A. \( \pi \)
• B. \( \dfrac{\pi}{\sqrt{2}} \)
• C. \( \dfrac{\pi}{\sqrt{3}} \)
• D. \( \dfrac{\pi}{2} \)

Hint:

For surfaces of revolution like cones, convert to cylindrical coordinates and use \( dS = \sqrt{1 + (dz/dr)^2} \, r \, dr \, d\theta. \)

Answer 1

The Correct Option is B

Step 1: Equation of cone. 
\( z = \sqrt{x^2 + y^2} \Rightarrow r = z \) in cylindrical coordinates.

Step 2: Limits of integration. 
\( z = 0 \) to \( z = 1. \)

Step 3: Surface element for \( z = f(r) \). 
\[ dS = \sqrt{1 + \left(\frac{\partial z}{\partial r}\right)^2} \, r \, d\theta \, dr. \] Since \( z = r, \frac{\partial z}{\partial r} = 1 \), \[ dS = \sqrt{2} \, r \, dr \, d\theta. \]

Step 4: Express \( x^2 + y^2 = r^2. \) 
\[ \iint_S (x^2 + y^2)\, dS = \int_0^{2\pi}\int_0^1 r^2 (\sqrt{2}r) \, dr\,d\theta = \sqrt{2}\int_0^{2\pi}\int_0^1 r^3 \, dr\,d\theta. \] \[ = \sqrt{2}(2\pi)\left[\frac{r^4}{4}\right]_0^1 = \frac{\pi}{\sqrt{2}}. \]

Final Answer: \( \dfrac{\pi}{\sqrt{2}}. \)