Question

Let S be the sphere of radius R.

Answer 1

Answer:

Step-by-step explanation:

a. Given that the sphere is symmetrical in nature. If we take a look at the yz-plane, over the hemisphere,  the integral on the surface of x on both sides of the plane will annul each other and the outcome will be zero. However, the remaining two integrals will also be zero as a result of symmetry on the xy and xz planes.

Thus, $\int \int_S \ x^2 dS = \int \int_S \ y^2 dS = \int \int_S \ z^2 dS$

b. Recall that :

$R^2=x^2+y^2+z^2$

By applying the integral of the surface area.

$\int \int_S x^2 \ dS + \int \int_S y^2 \ dS + \int \int_S z^2 \ dS = R^2 \int \int_S \ dS$

$R^2 \int \int_S \ dS = 4 \pi R^2$       (surface area of a sphere)

From above;

$\int \int_S \ x^2 dS = \int \int_S \ y^2 dS = \int \int_S \ z^2 dS$

∴

$\int \int_S \ x^2 dS = \int \int_S \ y^2 dS = \int \int_S \ z^2 dS = \dfrac{4}{3} \pi R^4$