Understanding the Area Bounded by g(y), y-axis, y=a, and y=b

When we talk about finding the area bounded by a function g(y), the y-axis, and horizontal lines y=a and y=b, we are essentially looking for the region enclosed by these boundaries in the yx-plane. This is a common topic in integral calculus, where we use definite integrals to find the area of such regions.

Key Concepts

• Function g(y): A function in terms of y, which defines the curve or boundary on one side of the region.
• y-axis: The vertical axis in the Cartesian coordinate system.
• y=a and y=b: Horizontal lines where 'a' and 'b' are constants, representing the lower and upper bounds of the region along the y-axis.

The Integral for Area

To find the area of the region bounded by the curve g(y), the y-axis, y=a, and y=b, we use the following integral:

$$ \text{Area} = \int_{a}^{b} g(y) \, dy $$

This integral sums up the infinitesimally small rectangles with height g(y) and width dy from y=a to y=b.

Table of Differences and Important Points

Aspect	Description
Direction of Integration	Integration is performed with respect to y, not x.
Limits of Integration	The lower limit is y=a, and the upper limit is y=b.
Function of Variable	The function is g(y), indicating that y is the independent variable.
Orientation of Area	The area is oriented vertically, between the y-axis and the curve g(y).

Formulas

The general formula for the area is:

$$ \text{Area} = \int_{a}^{b} g(y) \, dy $$

If g(y) intersects the y-axis within the interval [a, b], the area must be split into separate integrals at the points of intersection.

Examples

Let's go through a couple of examples to illustrate how to find the area bounded by g(y), the y-axis, y=a, and y=b.

Example 1: Area under a Linear Function

Consider the linear function g(y) = 2y + 3, and we want to find the area between this line, the y-axis, y=1, and y=4.

$$ \text{Area} = \int_{1}^{4} (2y + 3) \, dy $$

To solve the integral, we find the antiderivative of g(y):

$$ \int (2y + 3) \, dy = y^2 + 3y + C $$

Evaluating this antiderivative from y=1 to y=4:

$$ \text{Area} = [(4^2 + 3 \cdot 4) - (1^2 + 3 \cdot 1)] = (16 + 12) - (1 + 3) = 24 $$

Example 2: Area under a Quadratic Function

Consider the quadratic function g(y) = y^2, and we want to find the area between this parabola, the y-axis, y=0, and y=3.

$$ \text{Area} = \int_{0}^{3} y^2 \, dy $$

To solve the integral, we find the antiderivative of g(y):

$$ \int y^2 \, dy = \frac{1}{3}y^3 + C $$

Evaluating this antiderivative from y=0 to y=3:

$$ \text{Area} = \left[\frac{1}{3}(3^3) - \frac{1}{3}(0^3)\right] = \frac{1}{3}(27) = 9 $$

Conclusion

Finding the area bounded by g(y), the y-axis, y=a, and y=b involves setting up a definite integral with respect to y, using the function g(y) as the integrand, and evaluating the integral between the limits a and b. This method can be applied to any function g(y) to find the area of the region it encloses with the y-axis and the horizontal lines y=a and y=b.