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Dated: 30-10-2024

Exact Differential

If \(z = f(x, y, w)\) then

\[dz = \frac{\partial z}{\partial x}dx + \frac{\partial z}{\partial y}dy+ \frac{\partial z}{\partial w}dw\]

Any expression of the form \(dz = P dx + Q dy\) where \(P\) and \(Q\) are functions1 of \(x\) and \(y\) is an exact differential if it can be integrated2 to determine \(z\).

\[\because P = \frac{\partial z}{\partial x} \text{ and } Q = \frac{\partial z}{\partial y}\]
\[\frac{\partial P}{\partial y} = \frac{\partial z}{\partial y \partial x} \text{ and } \frac{\partial Q}{\partial z} = \frac{\partial z}{\partial x \partial y}\]
\[\because \frac{\partial z}{\partial x \partial y} = \frac{\partial z}{\partial y \partial x}\]
\[\therefore \frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x}\]

Example

\[dz = (3x^2 + 4y^2)dx + 8xy dy\]

Solution

\[P = 3x^2 + 4y^2\]
\[\frac{\partial P}{\partial y} = 8y\]
\[Q = 8xy\]
\[\frac{\partial Q}{\partial x} = 8y\]
\[\because \frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x}\]

Therefore, \(dz\) is the exact differential.

Integration2 Of Exact Differentials

\[dz = P dx + Q dy\]
\[\therefore z = \int P dx \text{ and } z = \int Q dy\]

Example

\[dz = (2xy + 6x) dx + (x^2 + 2y^3) dy\]
\[P = \frac{\partial z}{\partial x} = 2xy + 6x\]
\[z = \int (2xy + 6x) dx\]
\[z = x^2y + 3x^2 + f(y)\]

Here \(f(y)\) is the constant of integral.2

\[Q = \frac{\partial z}{\partial y} = x^2 + 2y^3\]
\[z = \int (x^2 + 2y^3) dy\]
\[z = x^2y + \frac{y^4}{2} + g(x)\]

We can find values of \(g(x)\) and \(f(y)\) such that both equations represent the same \(z\).
So by comparing both equations, we see that

\[f(y) = \frac{y^4} 2\]
\[g(x) = 3x^2\]
\[z = x^2y + 3x^2 + \frac{y^4}{2}\]

Area Enclosed by a Closed Curve

Imagine in a 2D plane,3 we have a curve, bounded by a function1 \(f(x)\) within the interval4 \([x_1, x_2]\).
Pasted image 20241016130628.png
We know that the area of this region is given by the integral2

\[A_1 = \int_{x_1}^{x_2} f(x) dx\]

Then there is another curve \(g(x)\) such that it intersects \(f(x)\) at 2 points at least, making a closed loop.
Similarly, the area under \(g(x)\) would be

\[A_2 = \int_{x_1}^{x_2} g(x) dx\]

Then the area of the closed loop is

\[\text{Area} = \int_{x_1}^{x_2} f(x) dx - \int_{x_1}^{x_2} g(x) dx\]

To make a closed loop with counter clockwise orientation,

\[\text{Area} = - \int_{x_2}^{x_1} f(x) dx - \int_{x_1}^{x_2} g(x) dx\]
\[\text{Area} = - \left(\int_{x_2}^{x_1} f(x) dx + \int_{x_1}^{x_2} g(x) dx \right)\]

Here, we are integrating2 \(g(x)\) from left to right and \(f(x)\) from right to left.

\[\text{Area} = - \oint y dx\]

Example

Determine the area within the interval4 \([0, 2]\) bounded by functions1 \(y = 4x\) and \(y = x^3\).

Solution

Pasted image 20241016131959.png

Let us define paths such that we integrate2 in a counter clockwise direction.

\[c_1 : y=x^3 \text{ from } x=0 \text{ to } 2\]
\[c_2 : y=4x \text{ from } x=2 \text{ to } 0\]
\[A = - \oint y dx\]
\[=-\left\{\int_{1}^{2}x^{3}dx+\int_{2}^{0}4xdx\right\}\]
\[=-\left\{\left(\frac{x^{4}}{4}\right)_{0}^{2}+\left(2x^{2}\right)_{0}^{2}\right\}\]
\[= 4\]

Example

Find area of triangle with vertices \(O(0, 0)\), \(A(5, 3)\) and \(B(2, 6)\).

Solution

Pasted image 20241016142057.png
The equations for the sides of the triangle are

\[m\overline{OA} = y = \frac 3 5x\]
\[m\overline{OB} = y = \frac 6 2x = 3x\]
\[m\overline{AB} = y = \frac {6 - 3}{2 - 5}x + 8 = 8 - \frac 3 3 x = 8 -x\]

Here \(8\) is the y intercept.5
Let us now define paths for our counter clockwise direction.

\[c_1: y = \frac 3 5 x \text{ from } x = 0 \text{ to } 5\]
\[c_2: y = 8 - x \text{ from } x = 5 \text{ to } 2\]
\[c_3: y = 3x \text{ from } x = 2 \text{ to } 0\]
\[\because A = - \oint y dx\]
\[=-\left\{\int_{0}^{5}\frac{3}{5}xdx+\int_{5}^{2}(8-x)dx+\int_{2}^{0}3xdx\right\}\]
\[= 12\]

Line Integral

Imagine a curve \(C\) from \(A\) and \(B\) and there is a particle at \(K\) on this curve.
Then imagine there is a force \(\vec{F}\) acting on the particle at \(K\)
This force can be broken down into its components such that one component is along the tangent vector6 of the curve and other one is along the normal to the curve.
Let us define \(\delta s\) along the curve which is the distance from \(K\) to \(L\)

\[\delta s = \lvert L - K \rvert\]

Then the work done in moving the particle on \(K\) is given by

\[\lim_{\delta s \to 0} \sum F_t \delta s = \int F_t ds \text{ from } A \text{ to } B\]
\[= \int_{AB} F_t ds\]

This integral2 is called line integral.
We can further break the tangent component \(F_t\) into its \(x\) component which is \(P dx\) and \(y\) component which is \(Q dy\).

References

Read more about notations and symbols.

  1. Read more about functions. ↩↩↩

  2. Read more about integrals. ↩↩↩↩↩↩↩

  3. Read more about planes. ↩

  4. Read more about intervals. ↩↩

  5. Read more about intercepts. ↩

  6. Read more about vectors. ↩