Question:
Let \( \alpha = \iint_S \vec{F} \cdot \hat{n} \, dS \), where \( \vec{F} = (2x + 3z)\hat{i} + (xz - y)\hat{j} + (y^2 + 2z)\hat{k} \) and \( S \) is the sphere with centre at \( (3, -1, 2) \) and radius \( 9 \). Here \( \hat{n} \) is the unit normal drawn outward and \( \hat{i}, \hat{j}, \hat{k} \) are unit vectors.
Then the value of \( \dfrac{1}{36\pi}\alpha \) is equal to ______. (Answer in integer)
Let \( \alpha = \iint_S \vec{F} \cdot \hat{n} \, dS \), where \( \vec{F} = (2x + 3z)\hat{i} + (xz - y)\hat{j} + (y^2 + 2z)\hat{k} \) and \( S \) is the sphere with centre at \( (3, -1, 2) \) and radius \( 9 \). Here \( \hat{n} \) is the unit normal drawn outward and \( \hat{i}, \hat{j}, \hat{k} \) are unit vectors.
Then the value of \( \dfrac{1}{36\pi}\alpha \) is equal to ______. (Answer in integer)
Then the value of \( \dfrac{1}{36\pi}\alpha \) is equal to ______. (Answer in integer)
Show Hint
Use the divergence theorem: for a closed surface, the flux equals the divergence integrated over the enclosed volume, and this divergence turns out to be constant here.
Updated On: Jul 21, 2026
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Verified By Collegedunia
Correct Answer: 81
Solution and Explanation
Step 1: Pick the right tool.
\( S \) is a closed surface, a full sphere, and \( \hat{n} \) points outward. Whenever a flux integral is taken over a closed surface, the Gauss divergence theorem turns it into a volume integral.
\[ \alpha = \iint_S \vec{F} \cdot \hat{n} \, dS = \iiint_V (\nabla \cdot \vec{F}) \, dV \]
where \( V \) is the solid ball enclosed by \( S \).
Step 2: Find the divergence of \( \vec{F} \).
\( \vec{F} = (2x+3z)\hat{i} + (xz-y)\hat{j} + (y^2+2z)\hat{k} \), so take the partial derivative of each component with respect to its own variable and add them.
\[ \nabla \cdot \vec{F} = \frac{\partial}{\partial x}(2x+3z) + \frac{\partial}{\partial y}(xz-y) + \frac{\partial}{\partial z}(y^2+2z) \]
\[ \nabla \cdot \vec{F} = 2 + (-1) + 2 = 3 \]
The divergence is a constant, \( 3 \), the same at every point in space.
Step 3: Turn the volume integral into 3 times the volume.
Since \( \nabla \cdot \vec{F} = 3 \) everywhere, it can be pulled out of the integral.
\[ \alpha = \iiint_V 3 \, dV = 3 \iiint_V dV = 3V \]
Here \( V \) is the volume of the ball of radius \( 9 \). The centre \( (3,-1,2) \) does not change the volume, only the radius matters.
Step 4: Compute the volume of the ball.
The volume of a ball of radius \( r \) is \( \dfrac{4}{3}\pi r^3 \). With \( r = 9 \),
\[ V = \frac{4}{3}\pi (9)^3 = \frac{4}{3}\pi (729) = 972\pi \]
Step 5: Find \( \alpha \) and then the required value.
\[ \alpha = 3V = 3(972\pi) = 2916\pi \]
Now divide by \( 36\pi \):
\[ \frac{1}{36\pi}\alpha = \frac{2916\pi}{36\pi} = 81 \]
Final Answer:
The value of \( \dfrac{1}{36\pi}\alpha \) is \( 81 \). \[ \boxed{81} \]
\( S \) is a closed surface, a full sphere, and \( \hat{n} \) points outward. Whenever a flux integral is taken over a closed surface, the Gauss divergence theorem turns it into a volume integral.
\[ \alpha = \iint_S \vec{F} \cdot \hat{n} \, dS = \iiint_V (\nabla \cdot \vec{F}) \, dV \]
where \( V \) is the solid ball enclosed by \( S \).
Step 2: Find the divergence of \( \vec{F} \).
\( \vec{F} = (2x+3z)\hat{i} + (xz-y)\hat{j} + (y^2+2z)\hat{k} \), so take the partial derivative of each component with respect to its own variable and add them.
\[ \nabla \cdot \vec{F} = \frac{\partial}{\partial x}(2x+3z) + \frac{\partial}{\partial y}(xz-y) + \frac{\partial}{\partial z}(y^2+2z) \]
\[ \nabla \cdot \vec{F} = 2 + (-1) + 2 = 3 \]
The divergence is a constant, \( 3 \), the same at every point in space.
Step 3: Turn the volume integral into 3 times the volume.
Since \( \nabla \cdot \vec{F} = 3 \) everywhere, it can be pulled out of the integral.
\[ \alpha = \iiint_V 3 \, dV = 3 \iiint_V dV = 3V \]
Here \( V \) is the volume of the ball of radius \( 9 \). The centre \( (3,-1,2) \) does not change the volume, only the radius matters.
Step 4: Compute the volume of the ball.
The volume of a ball of radius \( r \) is \( \dfrac{4}{3}\pi r^3 \). With \( r = 9 \),
\[ V = \frac{4}{3}\pi (9)^3 = \frac{4}{3}\pi (729) = 972\pi \]
Step 5: Find \( \alpha \) and then the required value.
\[ \alpha = 3V = 3(972\pi) = 2916\pi \]
Now divide by \( 36\pi \):
\[ \frac{1}{36\pi}\alpha = \frac{2916\pi}{36\pi} = 81 \]
Final Answer:
The value of \( \dfrac{1}{36\pi}\alpha \) is \( 81 \). \[ \boxed{81} \]
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