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 $\frac{1}{36\pi} \alpha$ is equal to ________. (answer in integer)
Using the Divergence Theorem
The problem involves calculating the flux $\alpha = \iint_S \vec{F} \cdot \hat{n} \, dS$ of a vector field $\vec{F}$ through a closed surface $S$, which is a sphere. The Divergence Theorem is the most efficient method here. It relates the surface integral (flux) to a volume integral of the divergence of the vector field.
The Divergence Theorem states: $ \alpha = \iint_S \vec{F} \cdot \hat{n} \, dS = \iiint_V (\nabla \cdot \vec{F}) \, dV $ where $V$ is the volume enclosed by the surface $S$.
Calculate Vector Field Divergence
Given the vector field $\vec{F} = (2x + 3z)\hat{i} + (xz - y)\hat{j} + (y^2 + 2z)\hat{k}$, we compute its divergence ($\nabla \cdot \vec{F}$): $ \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 $
Evaluate the Volume Integral
Substitute the divergence back into the Divergence Theorem equation: $ \alpha = \iiint_V 3 \, dV = 3 \iiint_V dV $ The term $\iiint_V dV$ is simply the volume of the region $V$.
The surface $S$ is a sphere with radius $R = 9$. The volume $V$ of a sphere is given by $V = \frac{4}{3}\pi R^3$. $ V = \frac{4}{3}\pi (9)^3 = \frac{4}{3}\pi (729) = 4\pi (243) = 972\pi $
Now, calculate $\alpha$: $ \alpha = 3 \times V = 3 \times (972\pi) = 2916\pi $
Compute the Final Value
The question asks for the value of $\frac{1}{36\pi} \alpha$. $ \frac{1}{36\pi} \alpha = \frac{1}{36\pi} (2916\pi) $ Cancel $\pi$ and perform the division: $ \frac{2916}{36} = 81 $
The value is 81.
Important Questions from Area Under Curve
- A function $y(x)$ is defined in the interval $[0, 1]$ on the x-axis as
$y(x) = \begin{cases} 2 & \text{if } 0 \le x < \frac{1}{3} \\ 3 & \text{if } \frac{1}{3} \le x < \frac{3}{4} \\ 1 & \text{if } \frac{3}{4} \le x \le 1 \end{cases}$
Which one of the following is the area under the curve for the interval $[0, 1]$ on the x-axis? - The area of the region bounded by the parabola $x = -y^2$ and the line $y = x + 2$ equals
The work done by the force $F = (x + y)\hat{i} - (x^2 + y^2)\hat{j}$, where $\hat{i}$ and $\hat{j}$ are unit vectors in $\vec{OX}$ and $\vec{OY}$ directions, respectively, along the upper half of the circle $x^2 + y^2 = 1$ from $(1,0)$ to $(-1,0)$ in the $xy$-plane is
In the figure shown above, PQRS is a square. The shaded portion is formed by the intersection of sectors of circles with radius equal to the side of the square and centers at S and Q.
The probability that any point picked randomly within the square falls in the shaded area is ___________.- If $f(x) = 2 \ln(\sqrt{e^x})$, what is the area bounded by $f(x)$ for the interval $[0, 2]$on the x-axis?
