Area Under Simple Curves

In elementary geometry, we learn formulas to calculate the areas of standard geometric figures like triangles, rectangles, trapezoids, and circles. However, these basic formulas fall short when we need to find the area enclosed by arbitrary curves or non-standard shapes.

Finding the area under simple curves, and the area bounded by lines, arcs of circles, parabolas, and ellipses. We do this by visualizing an area as the limit of a sum of infinitesimally thin rectangular strips.

To find the area bounded by the curve y = f(x), the x-axis, and the vertical lines (ordinates) x = a and x = b:

1. Imagine the area as being composed of a very large number of extremely thin vertical strips.

2. Consider an arbitrary elementary strip of height y and width dx.

3. The area of this elementary strip is \[dA = y \cdot dx\].

4. The total area A is the continuous sum (integral) of all these elementary areas from x = a to x = b.

\[A = \int_{a}^{b} y \, dx = \int_{a}^{b} f(x) \, dx\]

To find the area bounded by the curve x = g(y), the y-axis, and the horizontal lines y = c and y = d:

1. We consider horizontal strips instead of vertical ones.

2. The elementary strip has length x and infinitesimally small width dy.

3. The area of the strip is \[dA = x \cdot dy\].

\[A = \int_{c}^{d} x \, dy = \int_{c}^{d} g(y) \, dy\]

If a portion of the curve lies below the x-axis (i.e., f(x) < 0), evaluating the definite integral will yield a negative value. Since physical area cannot be negative, we must take the absolute value (modulus) of the integral.

\[\text{Area} = \left| \int_{a}^{b} f(x) \, dx \right|\]

If a curve crosses the x-axis within the interval [a, b], resulting in area \[A_1\] below the axis and area \[A_2\] above the axis, you cannot integrate from a to $b$ directly (as the negative and positive areas will cancel each other out). The total area is the sum of the absolute values of the separate areas.

\[\text{Total Area } A = |A_1| + A_2\]

Find the area enclosed by the ellipse \[\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\]

Solution:

The area of the region\[ABA'B'A\] bounded by the ellipse

\[= 4 \left( \text{area of the region AOBA in the first quadrant bounded by the curve, } x\text{-axis and the ordinates } x = 0, x = a \right)\]

(as the ellipse is symmetrical about both \[x\]-axis and \[y\]-axis)

\[= 4 \int_{0}^{a} y dx\] (taking vertical strips)

Now \[\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\] gives \[y = \pm \frac{b}{a} \sqrt{a^2 - x^2}\], but as the region AOBA lies in the first quadrant, \[y\] is taken as positive. So, the required area is