Determine the area enclosed by a curve, volume of a solid of revolution, and length of an arc

When two curves f(x) and g(x) bound a region, with f(x) ≥ g(x) on [a,b], the enclosed area is:

$A = \int_a^b [f(x) - g(x)] \, dx$

The limits a and b are found from the points of intersection of the two curves.

Example: Find the area between the curve y = (x-1)² and the line y = x + 1.

Solution

At intersection: $(x-1)^2 = x + 1$ $x^2 - 2x + 1 = x + 1$ $x^2 - 3x = 0$ $x(x - 3) = 0$ Thus, x = 0 and x = 3.

The area is: $A = \int_0^3 [(x + 1) - (x-1)^2] \, dx$ $A = \int_0^3 (3x - x^2) \, dx$ $A = \left[\frac{3x^2}{2} - \frac{x^3}{3}\right]_0^3 = \frac{9}{2} - 9 = -\frac{9}{2}$

Taking the absolute value: $A = \frac{9}{2} = 4.5 \text{ square units}$

For a curve y = f(x) from x = a to x = b, the arc length is given by:

$s = \int_a^b \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \, dx$

For parametric equations x = f(t), y = g(t): $s = \int_{t_1}^{t_2} \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} \, dt$

For polar coordinates r = f(θ): $s = \int_{\theta_1}^{\theta_2} \sqrt{r^2 + \left(\frac{dr}{d\theta}\right)^2} \, d\theta$

Example: Find the length of the arc of the curve y = 2x^(3/2) from x = 0 to x = 1/3.

Solution

Given y = 2x^(3/2), then: $\frac{dy}{dx} = 3\sqrt{x}$

The arc length is: $s = \int_0^{1/3} \sqrt{1 + 9x} \, dx$ $s = \left[\frac{2(1+9x)^{3/2}}{27}\right]_0^{1/3}$ $s = \frac{2}{27}[(1+9(\frac{1}{3}))^{3/2} - 1] = \frac{14}{27} \text{ units}$

When a region is rotated about an axis, the resulting solid's volume can be found using integration.

Disk Method (about x-axis): $V = \pi \int_a^b y^2 \, dx$

Washer Method (between two curves): $V = \pi \int_a^b \left[(f(x))^2 - (g(x))^2\right] \, dx$

Shell Method (about y-axis): $V = 2\pi \int_a^b x \cdot (\text{height}) \, dx$

Example: Find the volume generated when the region bounded by y = x² + 5, the x-axis, and x = 1 to x = 3 is rotated about the y-axis.

Solution

Using the shell method: $V = 2\pi \int_1^3 x(x^2 + 5) \, dx$ $V = 2\pi \int_1^3 (x^3 + 5x) \, dx$ $V = 2\pi \left[\frac{x^4}{4} + \frac{5x^2}{2}\right]_1^3$ $V = 2\pi\left[\left(\frac{81}{4} + \frac{45}{2}\right) - \left(\frac{1}{4} + \frac{5}{2}\right)\right]$ $V = 80\pi \text{ cubic units}$

In Tanzanian construction, understanding these integration concepts is practical for tasks like calculating the amount of concrete needed for curved architectural features. For instance, when building a domed roof or a cylindrical water tank, a builder uses volume calculation principles from integral calculus to determine how many bags of cement and cubic meters of sand are required—ensuring materials are ordered correctly and cost-effectively for the project.