Calculating Area Bounded by Curves in the First Quadrant

The problem asks for the area of the region enclosed by three curves in the first quadrant: \(x^2 + y^2 = \pi^2\), \(y = \sin x\), and \(x = 0\).

Let's analyze each curve:

• \(x = 0\): This is the equation of the y-axis.
• \(y = \sin x\): This is a trigonometric function. In the first quadrant, for \(0 \le x \le \pi\), the sine function starts at \(y=0\) (at \(x=0\)), increases to a maximum of \(y=1\) (at \(x=\pi/2\)), and decreases back to \(y=0\) (at \(x=\pi\)). The curve is always above or on the x-axis in this interval.
• \(x^2 + y^2 = \pi^2\): This is the equation of a circle centered at the origin \((0,0)\) with a radius of \(\pi\). In the first quadrant (\(x \ge 0\) and \(y \ge 0\)), the equation can be written as \(y = \sqrt{\pi^2 - x^2}\). This arc starts at \((0, \pi)\) and ends at \((\pi, 0)\).

We need to find the area of the region enclosed by these three boundaries in the first quadrant. Let's consider the region between \(x=0\) and \(x=\pi\).

At \(x=0\), \(y=\sin(0)=0\) and \(y=\sqrt{\pi^2-0^2}=\pi\). So the y-axis segment from \((0,0)\) to \((0,\pi)\) is part of the boundary defined by \(x=0\) and the circle arc \(x^2+y^2=\pi^2\).

At \(x=\pi\), \(y=\sin(\pi)=0\) and \(y=\sqrt{\pi^2-\pi^2}=0\). So both the sine curve and the circle arc meet at \((\pi,0)\) on the x-axis.

In the first quadrant, for \(0 \le x \le \pi\), the curve \(y=\sqrt{\pi^2-x^2}\) is always above or equal to the curve \(y=\sin x\). We can see this by checking points like \(x=0\) (\(\pi > 0\)), \(x=\pi/2\) (\(\sqrt{\pi^2-(\pi/2)^2} \approx 2.72 > 1\)), and \(x=\pi\) (\(0=0\)).

The region is bounded on the left by \(x=0\), from below by \(y=\sin x\), and from above by \(y=\sqrt{\pi^2 - x^2}\) up to \(x=\pi\). The area of this region can be calculated by integrating the difference between the upper curve (\(y=\sqrt{\pi^2-x^2}\)) and the lower curve (\(y=\sin x\)) with respect to \(x\) from the left boundary \(x=0\) to the right boundary \(x=\pi\).

The area \(A\) is given by the definite integral:

\(A = \int_{0}^{\pi} (\sqrt{\pi^2 - x^2} - \sin x) dx\)

We can split this into two separate integrals:

\(A = \int_{0}^{\pi} \sqrt{\pi^2 - x^2} dx - \int_{0}^{\pi} \sin x dx\)

Evaluating the First Integral: Area under the Circle Arc

The integral \(\int_{0}^{\pi} \sqrt{\pi^2 - x^2} dx\) represents the area under the curve \(y = \sqrt{\pi^2 - x^2}\) from \(x=0\) to \(x=\pi\). This curve in the first quadrant corresponds to the arc of the circle \(x^2+y^2=\pi^2\) from \((0,\pi)\) down to \((\pi,0)\). The area under this arc from \(x=0\) to \(x=\pi\) is exactly the area of the quarter circle of radius \(\pi\) in the first quadrant.

Geometrically, the area of a quarter circle with radius \(r\) is \(\frac{1}{4} \pi r^2\). Here, \(r = \pi\). So the area is \(\frac{1}{4} \pi (\pi)^2 = \frac{\pi^3}{4}\).

Alternatively, using integration formula \(\int \sqrt{a^2-x^2} dx = \frac{x}{2}\sqrt{a^2-x^2} + \frac{a^2}{2}\sin^{-1}(\frac{x}{a})\):

\(\int_{0}^{\pi} \sqrt{\pi^2 - x^2} dx = \left[ \frac{x}{2}\sqrt{\pi^2-x^2} + \frac{\pi^2}{2}\sin^{-1}\left(\frac{x}{\pi}\right) \right]_{0}^{\pi}\)

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

\(\frac{\pi}{2}\sqrt{\pi^2-\pi^2} + \frac{\pi^2}{2}\sin^{-1}\left(\frac{\pi}{\pi}\right) = \frac{\pi}{2}\sqrt{0} + \frac{\pi^2}{2}\sin^{-1}(1) = 0 + \frac{\pi^2}{2} \cdot \frac{\pi}{2} = \frac{\pi^3}{4}\)

Evaluate at the lower limit \(x=0\):

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

So, \(\int_{0}^{\pi} \sqrt{\pi^2 - x^2} dx = \frac{\pi^3}{4} - 0 = \frac{\pi^3}{4}\).

Evaluating the Second Integral: Area under the Sine Curve

The integral \(\int_{0}^{\pi} \sin x dx\) represents the area under the curve \(y = \sin x\) from \(x=0\) to \(x=\pi\).

\(\int_{0}^{\pi} \sin x dx = [-\cos x]_{0}^{\pi}\)

Evaluate at the limits:

\([-\cos x]_{0}^{\pi} = (-\cos \pi) - (-\cos 0) = (-(-1)) - (-1) = 1 + 1 = 2\)

Calculating the Total Area

Now, subtract the area under the lower curve from the area under the upper curve:

\(A = \int_{0}^{\pi} \sqrt{\pi^2 - x^2} dx - \int_{0}^{\pi} \sin x dx = \frac{\pi^3}{4} - 2\)

The area of the region enclosed in the first quadrant by \(x^2 + y^2 = \pi^2\), \(y = \sin x\) and \(x = 0\) is \(\frac{\pi^3}{4} - 2\).

Revision Table: Key Concepts

Additional Information: Understanding the Region of Area Calculation

Visualizing the region for area calculation is crucial. The region is in the first quadrant. The y-axis (\(x=0\)) forms the left boundary. The curve \(y=\sin x\) starts at \((0,0)\) and forms the lower boundary for \(0 \le x \le \pi\). The circle arc \(y=\sqrt{\pi^2-x^2}\) starts at \((0,\pi)\) on the y-axis and ends at \((\pi,0)\) on the x-axis, forming the upper boundary for \(0 \le x \le \pi\). Both the lower and upper boundary curves meet at \((\pi,0)\), which naturally defines the right limit of integration as \(x=\pi\). The region is literally the area 'sandwiched' between the circle arc above and the sine curve below, from \(x=0\) to \(x=\pi\).