Question

The surface integral \( \int_S x^2 \, dS \) over the upper hemisphere 
\[ z = \sqrt{1 - x^2 - y^2}  \]
with radius 1 is ..........

• A. \( \frac{\pi}{4} \)
• B. \( \frac{\pi}{3} \)
• C. \( \pi \)
• D. \( 2\pi \)

Hint:

When solving surface integrals over spherical surfaces, use spherical coordinates and remember that the surface element \( dS = \sin \theta \, d\theta \, d\phi \) and the integrand \( x^2 = \sin^2 \theta \cos^2 \phi \).

Answer 1

The Correct Option is A

We are given the surface integral \( \int_S x^2 \, dS \) over the upper hemisphere defined by the equation: \[ z = \sqrt{1 - x^2 - y^2} \] The surface is a hemisphere with radius 1. To calculate the surface integral, we use the formula for a surface integral over a parameterized surface: \[ \int_S f(x, y, z) \, dS = \iint_D f(x(u, v), y(u, v), z(u, v)) \left\| \frac{\partial(x, y, z)}{\partial(u, v)} \right\| \, du \, dv \] In spherical coordinates, for a hemisphere, the surface element \( dS \) is given by: \[ dS = \sin \theta \, d\theta \, d\phi \] where \( \theta \) is the polar angle, ranging from 0 to \( \frac{\pi}{2} \) (upper hemisphere), and \( \phi \) is the azimuthal angle, ranging from 0 to \( 2\pi \). Now, to compute the integral, we need the integrand \( x^2 \), which in spherical coordinates is \( x = \sin \theta \cos \phi \). Thus: \[ x^2 = \sin^2 \theta \cos^2 \phi \] The surface integral becomes: \[ \int_S x^2 \, dS = \int_0^{2\pi} \int_0^{\frac{\pi}{2}} \sin^2 \theta \cos^2 \phi \sin \theta \, d\theta \, d\phi \] First, evaluate the integral with respect to \( \phi \): \[ \int_0^{2\pi} \cos^2 \phi \, d\phi = \pi \] Next, integrate with respect to \( \theta \): \[ \int_0^{\frac{\pi}{2}} \sin^3 \theta \, d\theta = \frac{4}{3} \] Multiplying these results: \[ \int_S x^2 \, dS = \frac{4}{3} \times \pi = \frac{\pi}{4} \] Thus, the value of the surface integral is \( \frac{\pi}{4} \).