Question

The value of the surface integral $\iint_S (2x + z) dy dz + (2x + z) dz dx + (2z + y) dx dy$ over the sphere $S: x^2 + y^2 + z^2 = 9$ is

• A. $72\pi$
• B. $144\pi$
• C. $36\pi$
• D. $432\pi$

Hint:

When evaluating surface integrals over closed surfaces, the Divergence Theorem is often a simpler approach than direct integration. Remember to correctly compute the divergence of the vector field and the volume of the enclosed region.

Answer 1

The Correct Option is B

Let the vector field be $\mathbf{F} = (2x + z) \mathbf{i} + (2x + z) \mathbf{j} + (2z + y) \mathbf{k}$. The surface integral can be written as $\iint_S \mathbf{F} \cdot \mathbf{n} \, dS$. Since $S$ is a closed surface (a sphere), we can use the Divergence Theorem: $$\iint_S \mathbf{F} \cdot \mathbf{n} \, dS = \iiint_V (\nabla \cdot \mathbf{F}) \, dV$$ where $V$ is the volume enclosed by the sphere $x^2 + y^2 + z^2 = 9$. First, calculate the divergence of $\mathbf{F}$:
$$\nabla \cdot \mathbf{F} = \frac{\partial}{\partial x}(2x + z) + \frac{\partial}{\partial y}(2x + z) + \frac{\partial}{\partial z}(2z + y) = 2 + 0 + 2 = 4$$ Next, calculate the volume of the sphere $x^2 + y^2 + z^2 = 9$. The radius of the sphere is $R = \sqrt{9} = 3$. The volume of a sphere is given by $V = \frac{4}{3} \pi R^3$. $$V = \frac{4}{3} \pi (3)^3 = \frac{4}{3} \pi (27) = 36\pi$$ Now, evaluate the volume integral:
$$\iiint_V (\nabla \cdot \mathbf{F}) \, dV = \iiint_V 4 \, dV = 4 \iiint_V dV = 4 \times ({Volume of the sphere})$$ $$\iiint_V 4 \, dV = 4 \times 36\pi = 144\pi$$ Thus, the value of the surface integral is $144\pi$.