Question

For a given vector \(\vec{F}=-y\hat{i}+z\hat{j}+x^2\hat{k}\) , the surface integral \(\int_S(\vec{▽}\times \vec{F}).\hat{r}dS\) over the surface S of a hemisphere of radius R with the centre of the base at the origin is

• A. πR²
• B. \(\frac{2\pi R^2}{3}\)
• C. -πR²
• D. \(-\frac{2\pi R^2}{3}\)

Answer 1

The Correct Option is A

To solve the given problem, we need to find the surface integral \(\int_S (\vec{▽} \times \vec{F}) \cdot \hat{r} \, dS\) over the surface of a hemisphere of radius \(R\). The vector field is given by \(\vec{F} = -y\hat{i} + z\hat{j} + x^2\hat{k}\).

First, let's calculate the curl of the vector field \(\vec{F}\). The curl is given by the formula:

\[ \vec{▽} \times \vec{F} = \left| \begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ -y & z & x^2 \end{array} \right| \]

Calculating the determinant, we find:

\[ \vec{▽} \times \vec{F} = \hat{i}\left(\frac{\partial x^2}{\partial y} - \frac{\partial z}{\partial z}\right) - \hat{j}\left(\frac{\partial x^2}{\partial x} - \frac{\partial (-y)}{\partial z}\right) + \hat{k}\left(\frac{\partial z}{\partial x} - \frac{\partial (-y)}{\partial y}\right) \]

Simplifying further:

\[ \vec{▽} \times \vec{F} = \hat{i}(0 - 1) - \hat{j}(2x - 0) + \hat{k}(0 + 1) = -\hat{i} - 2x\hat{j} + \hat{k} \]

The hemisphere surface consists of the curved surface and the flat circular base. However, we only need to consider the surface integral over the curved surface, since the problem specifies surface \(S\).

For the curved surface of a hemisphere, we use the symmetry and the nature of the vector field. The normal to the surface at any point on the hemisphere is radial, pointing outward.

Using Stokes' Theorem, the surface integral of the curl of a vector field over a surface is equal to the line integral of the vector field over the boundary of that surface. Since there's no boundary on the curved surface apart from the circular base, and the vector field contribution to the base would be zero by symmetry, the integral simplifies just to that on the surface:

\[ \int_S (\vec{▽} \times \vec{F}) \cdot \hat{r} \, dS = \int_S 1 \, dS = \text{Surface area of hemisphere} \]

The surface area of a hemisphere is given by:

\[ 2\pi R^2 / 2 = \pi R^2 \]

Therefore, the value of the surface integral is \(\pi R^2\), which is the correct answer.