Problem 8 - Technical Mathematics with Calculus

Question

Find the first-quadrant area bounded by each curve and both coordinate axes. $$\sqrt{x}+\sqrt{y}=1$$

Step-by-Step Solution

Answer

The area of the first-quadrant region bounded by the curve \(\sqrt{x} + \sqrt{y} = 1\) and both coordinate axes is \(\frac{5}{6}\) square units.

Step 1: Identify the type of region

Recognize that the first quadrant is the region where both x and y are non-negative. The given equation describes a boundary of the region where the area needs to be found. To get the region bounded by the curve and both axes, we need to express y in terms of x or x in terms of y.

Step 2: Solve for y in terms of x

To express y in terms of x, isolate y on one side of the equation: \( \) \( \( y = (1 - \sqrt{x})^2 \) \), which is valid in the first quadrant (\( x \geq 0 \) and \( y \geq 0 \)).

Step 3: Sketch the region

Plot the curve \( \sqrt{x} + \sqrt{y} = 1 \) on a coordinate plane, noting where it crosses the axes. The region of interest is under this curve, and above the x-axis, and to the right of the y-axis.

Step 4: Set up the integral for area

To find the area under the curve, set up the integral with respect to x: \( \( \text{Area} = \int_{0}^{1} (1 - \sqrt{x})^2 \, dx \) \). The limits 0 and 1 come from the boundary of the curve when \( x = 0 \) and \( y = 0 \).

Step 5: Expand the integrand and integrate

Expand the integrand: \( (1 - \sqrt{x})^2 = 1 - 2\sqrt{x} + x \), and then integrate term by term: \( \int_{0}^{1} (1 - 2\sqrt{x} + x) \, dx = [x - 2/3 x^{3/2} + 1/2 x^2]_{0}^{1} \).

Step 6: Evaluate the integral

Evaluate the definite integral at the bounds 0 and 1: \( [1 - 2/3(1) + 1/2(1)] - [0 - 2/3(0) + 1/2(0)] = 1 - 2/3 + 1/2 \).

Step 7: Simplify the result

Simplify to find the area: \( 1 - 2/3 + 1/2 = 6/6 - 4/6 + 3/6 = 5/6 \). The area of the region is \( 5/6 \) square units.

Key Concepts

First Quadrant Bounded Area

The 'first quadrant bounded area' refers to a region enclosed by a curve or curves that lie entirely within the first quadrant of the Cartesian coordinate system. The first quadrant is where both the x and y coordinates of any point are positive.

Understanding the first quadrant bounded area is crucial in geometry and calculus, especially when we are interested in finding the space that a curve occupies. This area can take various shapes depending on the equation of the curve involved, but they will always be to the right of the y-axis and above the x-axis, given the non-negative values for x and y.

It is essential to correctly identify the region within the first quadrant because this determines the limits for our integral calculations. We must ensure the function is appropriately solved for in terms of a single variable, to apply the integration technique effectively for area calculation.

Region of Integration

The 'region of integration' in the context of finding the area under a curve or between curves is the specific portion on the graph for which we wish to calculate the area. It is bounded by the curves and the axes or other defined lines or curves.

Identifying this region correctly is a precursor to setting up integrals, which are the main tool used in calculating these areas. The boundaries of the region define the limits of integration, which are essential in performing definite integrals. In our exercise, the boundary was given by the equation \( \sqrt{x} + \sqrt{y} = 1 \) and the coordinate axes.

When sketching the region, we visualize where the curve intersects the axes, allowing us to understand better the behavior of the function and to set useful boundaries for calculating the area using integration.

Definite Integral Application

Definite integrals are a fundamental application of calculus used to calculate the area under a curve within a certain interval. This application is directly related to the concept of boundaries, where the area between the x-axis and the curve from one x value to another is sought.

In our scenario, once the region of integration is determined and the function is expressed in terms of a single variable, we calculate the area using a definite integral. The integral takes the general form \(\int_{a}^{b} f(x) \, dx\), where 'a' and 'b' are the lower and upper limits of integration, respectively, and 'f(x)' represents the function that defines the curve.

Integration computes the sum of infinitesimally small rectangles under the curve, producing the total area. Hence, a definite integral is not merely a theoretical concept, but a practical and effective tool in determining space under a curve over a specified region.

Curve Area Calculation

The process of 'curve area calculation' involves determining the size of the area beneath a curve and above the x-axis, within a given range of x-values. This process is pivotal in many fields, including physics, engineering, and economics.

Within the context of our exercise, the curve area calculation begins by expressing y in terms of x to facilitate a one-dimensional integral. Once the equation is reformulated, the area can be computed by setting up the definite integral with the determined boundaries, which stem from the region of integration. After the integrand is expanded and integrated term by term, the integral's evaluation from the lower to the upper limit provides us with the exact measurement of the area.

It's important for students to not only manipulate integrals algebraically but to also interpret their geometric meaning. This connection between abstract calculation and tangible geometry helps solidify the concepts and applications of integration.