Calculating the Area of a Region Bounded by Curves

We are asked to find the area of the region in the first quadrant bounded by the curves \(y = \sqrt{1−\text{x}^2}\), \(y = x\), and \(y = 0\).

Let's first understand what each equation represents:

• \(y = \sqrt{1−\text{x}^2}\): Squaring both sides gives \(y^2 = 1 - x^2\), which rearranges to \(x^2 + y^2 = 1\). This is the equation of a circle centered at the origin (0,0) with a radius of 1. Since \(y = \sqrt{1−\text{x}^2}\), we are considering the upper semi-circle. The restriction to the first quadrant means \(x \ge 0\) and \(y \ge 0\), so this is the part of the circle in the first quadrant.
• \(y = x\): This is a straight line passing through the origin with a slope of 1. In the first quadrant, it's the line segment from (0,0) upwards, making a 45-degree angle with the positive x-axis.
• \(y = 0\): This is the equation of the x-axis.

The region in question is in the first quadrant (\(x \ge 0, y \ge 0\)) and is enclosed by these three curves. Let's identify the points where these curves intersect in the first quadrant:

• Intersection of \(y=x\) and \(y=0\): \(x=0\), so the point is (0,0).
• Intersection of \(y=\sqrt{1-x^2}\) and \(y=0\): \(\sqrt{1-x^2} = 0 \implies 1-x^2 = 0 \implies x^2 = 1\). In the first quadrant, \(x=1\), so the point is (1,0).
• Intersection of \(y=x\) and \(y=\sqrt{1-x^2}\): \(x = \sqrt{1-x^2}\). Since we are in the first quadrant, \(x \ge 0\). Squaring both sides gives \(x^2 = 1-x^2 \implies 2x^2 = 1 \implies x^2 = \frac{1}{2}\). Since \(x \ge 0\), \(x = \frac{1}{\sqrt{2}}\). As \(y=x\), \(y = \frac{1}{\sqrt{2}}\). The intersection point is \((\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}})\).

The region is bounded by the x-axis from (0,0) to (1,0), the arc of the unit circle from (1,0) to \((\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}})\), and the line \(y=x\) from \((\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}})\) back to (0,0). This specific region can be interpreted geometrically as a sector of the unit circle.

Geometric Approach using Sector Area

The region is bounded by the lines passing through the origin (\(y=0\) and \(y=x\)) and the arc of the circle \(x^2+y^2=1\). This describes a sector of the unit circle.

The x-axis (\(y=0\)) corresponds to an angle of 0 radians with respect to the positive x-axis.

The line \(y=x\) makes an angle \(\theta\) with the positive x-axis such that \(\tan \theta = \text{slope} = 1\). In the first quadrant, this angle is \(\theta = \arctan(1) = \frac{\pi}{4}\) radians.

The region is the sector of the unit circle with radius \(r=1\) between the angles \(\theta_1 = 0\) and \(\theta_2 = \frac{\pi}{4}\). The central angle of this sector is \(\Delta\theta = \theta_2 - \theta_1 = \frac{\pi}{4} - 0 = \frac{\pi}{4}\).

The area of a sector of a circle with radius \(r\) and central angle \(\theta\) (in radians) is given by the formula:

Area = \(\frac{1}{2} r^2 \theta\)

Substituting \(r=1\) and \(\theta = \frac{\pi}{4}\):

Area = \(\frac{1}{2} (1)^2 \left(\frac{\pi}{4}\right) = \frac{1}{2} \cdot \frac{\pi}{4} = \frac{\pi}{8}\)

Calculus Approach using Integration

Alternatively, we can calculate the area using definite integrals. The region needs to be split into two parts based on the x-values where the upper boundary curve changes:

• Part 1: From \(x=0\) to \(x = \frac{1}{\sqrt{2}}\), the region is bounded above by \(y=x\) and below by \(y=0\).
• Part 2: From \(x = \frac{1}{\sqrt{2}}\) to \(x=1\), the region is bounded above by \(y=\sqrt{1-x^2}\) and below by \(y=0\).

The total area is the sum of the areas of these two parts:

Total Area = Area 1 + Area 2

Area 1 = \(\int_{0}^{1/\sqrt{2}} x \, dx\)

Area 2 = \(\int_{1/\sqrt{2}}^{1} \sqrt{1-x^2} \, dx\)

Calculating Area 1:

\(\int_{0}^{1/\sqrt{2}} x \, dx = \left[\frac{x^2}{2}\right]_{0}^{1/\sqrt{2}} = \frac{(1/\sqrt{2})^2}{2} - \frac{0^2}{2} = \frac{1/2}{2} - 0 = \frac{1}{4}\)

Calculating Area 2: This integral involves \(\sqrt{a^2-x^2}\) with \(a=1\). The standard integral formula is \(\int \sqrt{a^2 - x^2} \, dx = \frac{x}{2}\sqrt{a^2 - x^2} + \frac{a^2}{2}\arcsin\left(\frac{x}{a}\right)\).

\(\int \sqrt{1-x^2} \, dx = \frac{x}{2}\sqrt{1-x^2} + \frac{1}{2}\arcsin\left(x\right)\)

Now evaluate the definite integral:

Area 2 = \(\left[\frac{x}{2}\sqrt{1-x^2} + \frac{1}{2}\arcsin\left(x\right)\right]_{1/\sqrt{2}}^{1}\)

Evaluate at the upper limit \(x=1\):

\(\frac{1}{2}\sqrt{1-1^2} + \frac{1}{2}\arcsin(1) = \frac{1}{2}\sqrt{0} + \frac{1}{2}\left(\frac{\pi}{2}\right) = 0 + \frac{\pi}{4} = \frac{\pi}{4}\)

Evaluate at the lower limit \(x=1/\sqrt{2}\):

\(\frac{1/\sqrt{2}}{2}\sqrt{1-(1/\sqrt{2})^2} + \frac{1}{2}\arcsin\left(\frac{1}{\sqrt{2}}\right) = \frac{1}{2\sqrt{2}}\sqrt{1-\frac{1}{2}} + \frac{1}{2}\left(\frac{\pi}{4}\right) = \frac{1}{2\sqrt{2}}\sqrt{\frac{1}{2}} + \frac{\pi}{8} = \frac{1}{2\sqrt{2}} \cdot \frac{1}{\sqrt{2}} + \frac{\pi}{8} = \frac{1}{4} + \frac{\pi}{8}\)

Area 2 = (Value at \(x=1\)) - (Value at \(x=1/\sqrt{2}\))

Area 2 = \(\frac{\pi}{4} - \left(\frac{1}{4} + \frac{\pi}{8}\right) = \frac{\pi}{4} - \frac{1}{4} - \frac{\pi}{8} = \frac{2\pi - \pi}{8} - \frac{1}{4} = \frac{\pi}{8} - \frac{1}{4}\)

Total Area = Area 1 + Area 2 = \(\frac{1}{4} + \left(\frac{\pi}{8} - \frac{1}{4}\right) = \frac{1}{4} + \frac{\pi}{8} - \frac{1}{4} = \frac{\pi}{8}\)

Summary of Area Calculation

Both the geometric approach (treating the region as a sector of a circle) and the calculus approach (using integration) yield the same result for the area of the region bounded by \(y = \sqrt{1−\text{x}^2}\), \(y = x\), and \(y = 0\) in the first quadrant.

The calculated area is \(\frac{\pi}{8}\).

Revision Table: Area Bounded by Curves

Additional Information: Area Calculation Methods

Finding the area of a region bounded by curves is a common problem in calculus. There are typically two main approaches:

• Integration with respect to x: If the region is bounded above by a curve \(y=f(x)\) and below by \(y=g(x)\) from \(x=a\) to \(x=b\), the area is \(\int_a^b (f(x) - g(x)) \, dx\). If the lower boundary is the x-axis, \(g(x)=0\). For more complex regions, the area might need to be split into multiple integrals based on where the upper or lower boundary curves change.
• Integration with respect to y: If the region is bounded on the right by \(x=f(y)\) and on the left by \(x=g(y)\) from \(y=c\) to \(y=d\), the area is \(\int_c^d (f(y) - g(y)) \, dy\). If the left boundary is the y-axis, \(g(y)=0\).
• Geometric Formulas: For certain simple regions (like triangles, rectangles, circles, sectors), standard geometric formulas can be used. Sometimes, a complex region can be decomposed into simpler geometric shapes and/or areas calculated by integration. Recognizing geometric shapes can often simplify the problem significantly, as seen in this example where the region is a sector of a circle.

In this problem, the region is conveniently shaped like a sector of the unit circle, making the geometric method straightforward. The angle of the line \(y=x\) with the positive x-axis is key to determining the sector angle.

Remember to always sketch the region to understand the boundaries and determine the appropriate method (integration with respect to x or y) or if a geometric approach is possible.