Dated: 30-10-2024

Examples

Example

Evaluate the following integral¹ around the region \(R\) bounded by the curves \(y = x^2\) and \(x = y^2\) by using green's theorem.²

\[I = \oint_C (xy dx+ (2x - y) dy)\]

Solution

First of all, for the boundaries of integral,¹ we need to find the intersection points.

\[\because y^2 = x\]

\[\therefore y = \sqrt x\]

We will ignore the \(-\sqrt x\) because we only care about first quadrant.³
We know that the \(y\) values will be the same for both equations at the intersections.
Therefore,

\[\sqrt x = x^2\]

Squaring both sides, we get

\[x = x^4\]

\[0 = x^4 - x\]

\[0 = x(x^3 - 1)\]

\[x = 0 \text{ and } x^3 - 1 = 0\]

\[\implies x^3 = 1\]

\[x = 1\]

We can put \(x = 0, 1\) in either equation and we will get \(y = 0, 1\).
Therefore, our intersection points are \((0, 0)\) and \((1, 1)\).

\[\because I = \oint_C (xy dx+ (2x - y) dy)\]

\[\therefore P = xy \implies \frac{\partial P}{\partial y} = x\]

\[\therefore Q = 2x - y \implies \frac{\partial Q}{\partial x} = 2\]

Using the green's theorem²

\[= - \iint_R \left(\frac{\partial P}{\partial y} - \frac{\partial Q}{\partial x}\right)dxdy\]

\[= - \iint_R (x - 2)dxdy\]

\[= - \int_0^1\int_{y = x^2}^{y = \sqrt x} (x - 2) dy dx\]

\[= - \int_0^1 (x - 2) \left[y\right]_{x^2}^{\sqrt x}dx\]

\[=-\int_{0}^{1}(x-2)(\sqrt{x}-x^{2})dx\]

\[=\int_{0}^{1}(x^{\frac 3 2}-x^{3}-2x^{\frac 1 2}+2x^{2})dx\]

\[=-\left[\frac{2}{5}x^{\frac 5 2}-\frac{1}{4}x^{4}-\frac{4}{3}x^{\frac 3 2}+\frac{2}{3}x^{3}\right]_{0}^{1}\]

\[= \frac {31}{60}\]

Gradient of a Scalar `Function`⁴

If \(\phi\) is a scalar function⁴ then its gradient is

\[\nabla \phi = \left(\hat i \frac {\partial}{\partial x} + \hat j \frac {\partial}{\partial y} + \hat k \frac {\partial}{\partial z}\right) \phi\]

\[= \hat i \frac {\partial \phi}{\partial x} + \hat j \frac {\partial \phi}{\partial y} + \hat k \frac {\partial \phi}{\partial z}\]

Divergence of a Vector Function

If \(\vec A(a_1, a_2, a_3) = a_1 \hat i + a_2 \hat j + a_3 \hat k\)
then

\[\nabla \cdot \vec A=\left(\hat{i}\frac{\partial}{\partial x}+\hat{j}\frac{\partial}{\partial y}+\hat{k}\frac{\partial}{\partial z}\right)\cdot(\hat{a}_{1}\hat{i}+\hat{a}_{2}\hat{j}+\hat{a}_{3}\hat{k})\]

\[=\frac{\partial a_{1}}{\partial x}+\frac{\partial a_{2}}{\partial y}+\frac{\partial a_{3}}{\partial z}\]

Curl of a Vector Function

If \(\vec A(a_1, a_2, a_3) = a_1 \hat i + a_2 \hat j + a_3 \hat k\)
then

\[\nabla \times \vec A=\left(\hat{i}\frac{\partial}{\partial x}+\hat{j}\frac{\partial}{\partial y}+\hat{k}\frac{\partial}{\partial z}\right)\times(\hat{a}_{1}\hat{i}+\hat{a}_{2}\hat{j}+\hat{a}_{3}\hat{k})\]

\[ = \left| \begin{matrix} \hat i & \hat j & \hat k \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ a_1 & a_2 & a_3 \end{matrix} \right| \]

\[\nabla\times \vec A=\hat{i}\left(\frac{\partial a_{3}}{\partial y}-\frac{\partial a_{2}}{\partial z}\right)+\hat{j}\left(\frac{\partial a_{1}}{\partial z}-\frac{\partial a_{3}}{\partial x}\right)+\hat{k}\left(\frac{\partial a_{2}}{\partial x}-\frac{\partial a_{1}}{\partial y}\right)\]

After playing around with these, you will come across following results

\[\nabla \times (\nabla \phi) = 0\]
\[\nabla \cdot (\nabla \times \vec A) = 0\]
\[\nabla \cdot (\nabla \phi) = \frac{\partial^{2}\phi}{\partial x^{2}}+\frac{\partial^{2}\phi}{\partial y^{2}}+\frac{\partial^{2}\phi}{\partial z^{2}}\]