Question

The value of \(\iint_S (x^2+2y^2+3z^2)\,ds\) over the sphere \(x^2+y^2+z^2=9\) is

• A. \(648\pi\)
• B. \(324\pi\)
• C. \(216\pi\)
• D. \(108\pi\)

Hint:

On a sphere, use symmetry: \(\iint x^2ds=\iint y^2ds=\iint z^2ds\).

Answer 1

The Correct Option is A

Concept:
For a sphere \[ x^2+y^2+z^2=a^2 \] by symmetry, \[ \iint_S x^2\,ds=\iint_S y^2\,ds=\iint_S z^2\,ds \] Also, \[ x^2+y^2+z^2=a^2 \] on the sphere.

Step 1: Identify radius. Given, \[ x^2+y^2+z^2=9 \] So, \[ a=3 \] Surface area of sphere is \[ 4\pi a^2=4\pi(3)^2=36\pi \]

Step 2: Integrate \(x^2+y^2+z^2\). On the sphere, \[ x^2+y^2+z^2=9 \] Therefore, \[ \iint_S (x^2+y^2+z^2)\,ds = \iint_S 9\,ds \] \[ =9(36\pi) \] \[ =324\pi \]

Step 3: Use symmetry. Since each of \[ \iint_S x^2\,ds,\quad \iint_S y^2\,ds,\quad \iint_S z^2\,ds \] is equal, \[ \iint_S x^2\,ds = \iint_S y^2\,ds = \iint_S z^2\,ds = \frac{324\pi}{3} \] \[ =108\pi \]

Step 4: Evaluate the required integral. \[ \iint_S (x^2+2y^2+3z^2)\,ds \] \[ = \iint_S x^2\,ds + 2\iint_S y^2\,ds + 3\iint_S z^2\,ds \] \[ =108\pi+2(108\pi)+3(108\pi) \] \[ =(1+2+3)108\pi \] \[ =6(108\pi) \] \[ =648\pi \]

Step 5: Final answer. \[ \boxed{648\pi} \]