Dated: 30-10-2024

Uses of `Integrals`¹

Area as `Integral`¹

Imagine a triangle with \(base = 4\) on x axis in interval² \([0, 4]\) and \(height = 8\).
The hypotenuse is represented by \(y = 2x\).

\[\int_0^4 2x dx = \left[x^2\right]_0^4\]

\[= 4^2 - 0^2\]

\[= 16\]

We can also use the normal formula

\[\text{Area} = \frac 1 2 \text{base} \times \text{height}\]

\[= \frac 1 2 (4) \times (8) = 16\]

Volume as `Integral`¹

Similarly, we can use \(\iint\) to find volume.

Properties

\[\iint_R cf(x, y) dxdy = c\iint_R f(x, y) dxdy \text{ where } c \text{ is constant}\]
\[\iint_R (f(x, y) + g(x, y)) dxdy = \iint_R f(x, y) dxdy + \iint_R g(x, y) dxdy\]
\[\iint_R (f(x, y) - g(x, y)) dxdy = \iint_R f(x, y) dxdy - \iint_R g(x, y) dxdy\]
\[\iint_R f(x, y) dxdy \ge 0 \text{ if } f(x, y) \ge 0 \text{ on } R\]
\[\iint_R f(x, y) dxdy \ge \iint_R g(x, y) dxdy \text{ if } f(x, y) \ge g(x, y)\]

If \(f(x, y)\) is a non negative function³ defined over a region \(R\) then sub dividing \(R\) into \(R_1\) and \(R_2\) has effect of dividing the solid bounded by \(R\) and \(z = f(x, y)\) into 2 solids.

\[\iint_R f(x, y) dxdy = \iint_{R_1} f(x, y) dxdy + \iint_{R_2}f(x, y) dxdy\]

Computing Cross Section Area

Let's say we have \(f(x, y)\) bounded within \(a \le x \le b\) and \(c \le y \le d\).
Then if we take a certain value of \(y\), the \(f(x, y_0)\) becomes a function³ of \(x\).
Thus integrating it gives us area under the curve \(f(x, y_0)\) in interval² \(a \le x \le b\).
Therefore,

\[A(y) = \int_a^b f(x, y) dx\]

Double `Integral`¹ for Non Rectangular Region

Let's say we have a region \(R\) bounded by lines⁴ \(x = a\), \(x = b\), \(y = g_1(x)\) and \(y = g_2(x)\).
Therefore, our boundaries are \(a \le x \le b\) and \(g_1(x) \le y \le g_2(x)\).
This gives us volume as

\[V = \int_a^b \int_{g_1(x)}^{g_2(x)} f(x, y) dydx\]

Uses of Integrals1

Area as Integral1

Volume as Integral1