Question

The area of the region bounded by the circle \(x^{2}+y^{2}=12\) and parabola \(x^{2}=y\) is \(\ldots \ldots \ldots\) (a) \((2 \pi-\sqrt{3})\) Sq. unit (b) \(4 \pi+\sqrt{3}\) Sq. unit (c) \(2 \pi+\sqrt{3}\) Sq. unit (d) \(\pi+(\sqrt{3} / 2)\) Sq. unit

Short answer

The area of the region bounded by the circle \(x^2 + y^2 = 12\) and parabola \(x^2 = y\) is 24 Sq. units.

Step-by-step solution

Find the points of intersection between the circle and the parabola

To find the points where the circle and the parabola intersect, we can set \(x^2\) in the parabola equation equal to \(y\) in the circle equation and solve for x: \(x^2 + x^2 = 12\) Solve for x: \(2x^2 = 12\) \(x^2 = 6\) \(x=\pm\sqrt{6}\) Now that we have the x-coordinates of the points of intersection, we can find the y-coordinates by plugging these values back into the equation for the parabola: \(y = x^2\) \(y = (\sqrt{6})^2\) \(y = 6\) So, the points of intersection are \((\sqrt{6}, 6)\) and \((- \sqrt{6}, 6)\).

Set up the integral to find the area

Now that we have the points of intersection, we can set up an integral to find the area between the curves. We will use the formula for the area of the region bounded by two curves in a given interval [a, b]: \(\int_{a}^{b} (f(x)-g(x)) dx\) Here, \(f(x) = x^2\) is the parabola function (the upper curve in the region), and \(g(x) = \sqrt{12 - x^2}\) is the circle function (the lower curve in the region) after solving the circle equation for y. The interval [a, b] is given by the x-coordinates of the intersection points [-\(\sqrt{6}\), \(\sqrt{6}\)].

Calculate the integral

Now we can substitute our functions and interval into the area formula: \(\int_{-{\sqrt{6}}}^{\sqrt{6}} [(x^2) - (\sqrt{12 - x^2})] dx\) To solve this integral, we can notice that the integrand is an odd function, thus the integral should be equal to 0. However, we can simply take the integral from 0 to \(\sqrt{6}\) and multiply it by 2. We have: \[ 2\int_{0}^{\sqrt{6}} [(x^2) - (\sqrt{12 - x^2})] dx \] We will now solve the integral using the substitution method. Let \(u = 12 - x^2\), so \(du = -2x dx\): \[ -2\int [6 - (\sqrt{u})] du \] Now we can solve the integral: \(-2 [6u - \frac{2}{3}u^{\frac{3}{2}}]_{0}^{6}\) Evaluate the integral at the bounds: \(-2 [6(6) - \frac{2}{3}(6)^{\frac{3}{2}}] + 1 [6(0) - \frac{2}{3}(0)^{\frac{3}{2}}]\) Calculate the result: \(-2 [36 - 4\sqrt{6^3}] = -2 [36 - 48]\) \(-72 + 96 = 24\) The area of the region bounded by the circle and the parabola is 24 Sq. unit, which is not among the given choices. Thus, there may be an error in the problem or the choices provided.

Key concepts

Circle Equation

The equation of a circle in the standard form is given as: \( x^2 + y^2 = r^2 \). Here, \( r \) is the radius of the circle. This is a simple representation where every point \((x, y)\) on the circle maintains the distance \( r \) from the center, which is located at the origin (0,0) if no shifts are applied.

In this problem, the circle equation \( x^2 + y^2 = 12 \) indicates that the circle has a radius \( r \) such that \( r^2 = 12 \), or \( r = \sqrt{12} = 2\sqrt{3} \). Understanding how changes to the equation affect the circle, such as transformations which move the circle around the plane, is key to integral calculus solutions involving circles.

Parabola Equation

Parabolas are defined by their key equation, usually in the format \( y = ax^2 \) or \( x = ay^2 \), depending on their orientation. In this exercise, the given equation \( x^2 = y \) implies a parabola opening upwards.

The coefficient of \( x^2 \) determines the extent of the parabola's width. If this coefficient changes, the parabola gets wider or narrower. Understanding parabolas in this form involves recognizing key points like

• the vertex,
• the axis of symmetry,
• and how they intersect with other curves.

In our exercise, the parabola intersects at two symmetrical points along the x-axis with the circle, guiding us in setting up our integral for calculating the area between them.

Integral Calculus

Integral calculus is a fundamental concept for finding areas between curves. It involves calculating the definite integral of the difference between two function values over a defined interval.

In this context, the area between the circle and parabolic curves is calculated using the integral:

• \( \int_{a}^{b} (f(x) - g(x)) \, dx \),

where \( f(x) \) and \( g(x) \) are the equations of the curves (parabola and circle, respectively). Calculating this integral helps us determine the region on the graph between these intersecting curves.

The core of the solution lies in substituting correctly derived functions and solving the integral with respect to the intersection limits \( [-\sqrt{6}, \sqrt{6}] \). This shows us the enclosed "excess" area between the curves, showcasing integral calculus's capability to measure complex shapes.

Points of Intersection

The intersection points are critical when studying areas between curves. They represent the boundaries where two equations \( x^2 = y \) (parabola) and \( x^2 + y^2 = 12 \) (circle) meet.

To find these points, we need to solve simultaneous equations to determine \((x,y)\) pairs:

• By substituting and solving: \( x^2 + x^2 = 12 \), we get \( x^2 = 6 \),
• leading to the x-values \( \pm \sqrt{6} \).

Substitute these back into the parabola equation to confirm both yield \( y = 6 \). Thus, the points \( (\sqrt{6}, 6) \) and \( (- \sqrt{6}, 6) \) define our area of interest.

Recognizing these intersections guides us in setting the limits of integration, vital for calculating the area between these complex forms accurately.