Question

The area of the region bounded by curves \(x^{2}+y^{2}=4, x=1\) \(\& x=\sqrt{3}\) is..... (a) \((\pi / 3)\) sq. unit (b) \((2 \pi / 3)\) sq. unit (c) \((5 \pi / 6)\) sq. unit (d) \((4 \pi / 3)\) sq. unit

Short answer

The area of the region bounded by the curves \(x^2 + y^2 = 4\), \(x = 1\), and \(x = \sqrt{3}\) is \(\boxed{(d) \frac{4\pi}{3}\text{ sq. unit}}\).

Step-by-step solution

Find intersection points of the curves

We are given two curves: \(x^2 + y^2 = 4\) (which is a circle of radius 2 centered at the origin) and the vertical lines \(x = 1\) and \(x = \sqrt{3}\). These vertical lines will intersect the circle at different points since the circle's equation is symmetric with respect to both axes. For each vertical line, substitute the given x-value into the circle's equation to find the corresponding y-values: For \(x = 1\), we have: \(1^2 + y^2 = 4\) \(y^2 = 3\) \(y = \pm\sqrt{3}\) For \(x = \sqrt{3}\), we have: \((\sqrt{3})^2 + y^2 = 4\) \(y^2 = 1\) \(y = \pm 1\) So, the intersection points are \((1, \sqrt{3})\), \((1, -\sqrt{3})\), \((\sqrt{3}, 1)\), and \((\sqrt{3}, -1)\).

Set up integrals to find the area

Let's determine the area enclosed by the two curves and the lines \(x=1\) and \(x=\sqrt{3}\) using integration. Since we know the intersections points, we can set up two separate integrals, one for each side of the circle (above and below the x-axis), and then multiply the result by 2: The integrand will be the function for the upper half of the circle minus the lower half of the circle. The equation of the circle is \(x^2 + y^2 = 4\), so \(y^2 = 4 - x^2\), and \(y = \pm\sqrt{4 - x^2}\). So, the area enclosed by the curve between \(x = 1\) and \(x = \sqrt{3}\) is: \(A = 2 \int_{1}^{\sqrt{3}} (\sqrt{4 - x^2} - (-\sqrt{4 - x^2}))dx\)

Evaluate the integral

Now, we will evaluate the integral: \(A = 2 \int_{1}^{\sqrt{3}} (2\sqrt{4 - x^2})dx\) We can use substitution to solve this integral. Let \(x = 2\sin{u}\), then \(dx = 2\cos{u}du\), and the integral becomes: \(A = 2 \int_{\arcsin{1/2}}^{\arcsin{\sqrt{3}/2}} (2\sqrt{4 - 4\sin^2{u}})(2\cos{u})du\) Simplifying and using the identity \(\sqrt{1 - \sin^2{u}} = \cos{u}\): \(A = 8 \int_{\arcsin{1/2}}^{\arcsin{\sqrt{3}/2}} \cos^2{u} du\) Now we can apply the double-angle formula for \(\cos^2{u}\): \(\cos^2{u} = \frac{1 + \cos{2u}}{2}\) Now the integral becomes: \(A = 8 \int_{\arcsin{1/2}}^{\arcsin{\sqrt{3}/2}} \frac{1 + \cos{2u}}{2} du\) \(A = 8 \int_{\arcsin{1/2}}^{\arcsin{\sqrt{3}/2}} \frac{1}{2}du + 4 \int_{\arcsin{1/2}}^{\arcsin{\sqrt{3}/2}} \cos{2u} du\) Evaluating those integrals: \(A = 4(\arcsin{\sqrt{3}/2} - \arcsin{1/2}) + 4\left[\frac{1}{2}\sin{2u}\right]_{\arcsin{1/2}}^{\arcsin{\sqrt{3}/2}}\) Now plug in the values and simplify the expression: \(A = 4\left(\frac{\pi}{3} - \frac{\pi}{6}\right) + 4\left[\frac{1}{2}\sin{\left(4\cdot\frac{\pi}{3}\right)} - \frac{1}{2}\sin{\left(4\cdot\frac{\pi}{6}\right)}\right]\) \(A = 4\cdot\frac{\pi}{6} + 4\left[\frac{1}{2}(-\sqrt{3}/2) - \frac{1}{2}(1)\right]\) \(A = \frac{4\pi}{3}\) sq. unit So, the area of the region bounded by the curves is \((d) \frac{4\pi}{3}\) sq. unit.

Key concepts

Integration by Substitution

Integration by substitution is a powerful technique to simplify complex integrals by transforming them into more manageable forms. It involves changing variables to make integration easier.

Here’s how it works:

• Choose a substitution: Pick a variable change that simplifies the integral. For trigonometric functions, using base angles like sine or cosine helps a lot.
• Rewrite the integral: Transform the integral into a new one using the substitution.
• Adjust limits: If definite limits are given, convert these limits according to the new variable.
• Solve the transformed integral: Once simplified, integrate using basic formulas or rules.
• Back-substitute: Replace the variable with the original to get the final expression.

In this exercise, we used the substitution \( x = 2\sin{u} \) to evaluate the integral, changing a complex integral into a more accessible form. This technique is especially useful when dealing with square roots and trigonometric expressions.

Area Bounded by Curves

Finding the area between curves involves integrating the space between two functions. This can be visualized as taking one curve and subtracting the area under a second curve.

Here are the key steps:

• Identify boundaries: Establish the limits of integration where the curves intersect or are explicitly bounded.
• Set up the integral: The area is found by integrating the top function minus the bottom function between the limits.
• Calculate: Once the integral is set up, calculate it to find the area.

This method is especially useful in visualizing and analyzing geometric shapes. In our problem, we compute the area trapped between portions of a circle and straight lines. This area represents a slice of the circle.

Trigonometric Integration

Trigonometric integration handles integrals involving trigonometric functions. These techniques are necessary when dealing with circles or periodic functions.

Common strategies include:

• Use identities: Employ trigonometric identities to simplify expressions, like converting \( \cos^2{u} \) using the double-angle formula \( \cos^2{u} = \frac{1 + \cos{2u}}{2} \).
• Substitution: Change variables to turn trigonometric functions into simpler algebraic ones.
• Break down the integral: Sometimes splitting the integral helps in managing complex expressions more effectively.

In our exercise, trigonometric functions simplified the integrals derived from the circle equation, allowing us to solve them efficiently.

Circle Equations

Circle equations define circular shapes on a plane and are a cornerstone of geometry. The standard equation of a circle centered at the origin is \( x^2 + y^2 = r^2 \), where \( r \) is the radius.

Here’s how they work:

• Center and radius: The equation \( x^2 + y^2 = r^2 \) describes a circle with a center at (0,0) and a radius \( r \).
• Symmetry: Circles are symmetric about both the x-axis and y-axis, hence any vertical line will intersect the circle twice.
• Applications: These equations are used in problems involving geometry, trigonometry, and complex numbers.

In this exercise, the circle equation \( x^2 + y^2 = 4 \) illustrates using geometric properties to solve for areas, intersections, and bounds between curves.