Problem 31
Question
The area of the portion of the circle \(x^{2}+y^{2}=1\), which lies inside the parabola \(y^{2}=1-x\), is (A) \(\frac{\pi}{2}-\frac{2}{3}\) (B) \(\frac{\pi}{2}+\frac{2}{3}\) (C) \(\frac{\pi}{2}+\frac{4}{3}\) (D) \(\frac{\pi}{2}-\frac{4}{3}\)
Step-by-Step Solution
Verified Answer
The correct answer is (D) \(\frac{\pi}{2} - \frac{4}{3}\).
1Step 1: Identify Points of Intersection
To determine where the circle \(x^2 + y^2 = 1\) intersects the parabola \(y^2 = 1 - x\), equate \(x^2 + y^2 = 1\) and substitute \(y^2 = 1 - x\) into the circle equation. Substituting gives \(x^2 + 1 - x = 1\), simplifying to \(x^2 - x = 0\). Factoring yields \(x(x-1) = 0\), which gives \(x = 0\) or \(x = 1\). Substituting these into \(y^2 = 1 - x\), the points are \((0, \pm 1)\) and \((1, 0)\).
2Step 2: Find Relevant Area Boundaries
The parabola \(y^2 = 1-x\) opens to the left and the circle \(x^2 + y^2 = 1\) extends from the origin. The intersection points form a boundary for integration. We need the area where \(x^2 + y^2 = 1\) lies inside \(y^2 = 1-x\), specifically between \(x = 0\) and \(x = 1\).
3Step 3: Setup Integral for Circle Area
The area of the circle from \(x=0\) to \(x=1\) can be found by setting up the integral \[A_1 = 2\int_0^1 \sqrt{1-x^2} \; dx\], where \(2\) accounts for symmetry along the x-axis. This integral calculates the upper half of the circle under \(y^2 = 1 - x\).
4Step 4: Setup Integral for Parabola Area
Similarly, find the area for the parabola using the integral \[A_2 = 2\int_0^1 \sqrt{1-x} \; dx\]. Again, the factor of \(2\) accounts for symmetry. This integral captures the region of the parabola needed to subtract from the circle area.
5Step 5: Evaluate the Integrals
The integral for the semicircle is \(A_1 = 2\int_0^1 \sqrt{1-x^2} \; dx\) which evaluates to \(\frac{\pi}{4}\). The integral for the parabola is \(A_2 = 2\int_0^1 \sqrt{1-x} \; dx\), which simplifies to \(\frac{4}{3}\).
6Step 6: Calculate the Area Inside Both Curves
The area of the portion of the circle inside the parabola is obtained by subtracting the area under the parabola from the circle: \(\text{Area} = A_1 - A_2 = \frac{\pi}{2} - \frac{4}{3}\).
7Step 7: Identify the Correct Answer
Based on the calculated integral result, the area of the circle portion inside the parabola is \(\frac{\pi}{2} - \frac{4}{3}\). This corresponds to answer choice (D).
Key Concepts
Area of regionsIntersection of curves
Area of regions
In mathematical problems involving integration, one common task is determining the area of regions bounded by various curves. When calculating these areas, understanding the boundaries and limits is crucial.
Consider the area of a region enclosed within curves, such as a circle and a parabola. The main goal is to find the total space inside the circle that overlaps with the parabola. To achieve this, we utilize definite integrals to sum up the infinitesimally small areas of vertical strips that lie between the curves.
Integration helps us pinpoint the precise amount of space a region covers by adding together slices of space along a line. Hence, it becomes essential in solving complex problems involving overlapping regions.
Consider the area of a region enclosed within curves, such as a circle and a parabola. The main goal is to find the total space inside the circle that overlaps with the parabola. To achieve this, we utilize definite integrals to sum up the infinitesimally small areas of vertical strips that lie between the curves.
- First, identify the points where the curves intersect. These points establish the range for our integrals.
- Then, compute the area under each curve within the common interval using integration.
- Finally, subtract the area under the parabola from the area under the circle to find the area of the region common to both.
Integration helps us pinpoint the precise amount of space a region covers by adding together slices of space along a line. Hence, it becomes essential in solving complex problems involving overlapping regions.
Intersection of curves
The intersection of curves is a fundamental concept in mathematics, especially in calculus and algebra. Intersections pinpoint where two curves meet, which is crucial in various applications. Focusing on regions bounded by a circle and a parabola, it becomes essential to find where these two shapes cross each other.
To find intersection points:
To find intersection points:
- Set the equations of curves equal to each other. For instance, equate the equation of a circle like \(x^2 + y^2 = 1\) with that of a parabola such as \(y^2 = 1 - x\).
- Solve the resulting equations to find the \
Other exercises in this chapter
Problem 29
The area bounded by the lines \(y=2, x=1, x=a\) and the curve \(y=f(x)\), which cuts the last two lines above the first line for all \(a \geq 1\), is equal to \
View solution Problem 30
The area above \(x\)-axis, bounded by the line \(x=4\) and the curve \(y=f(x)\), where \(f(x)=x^{2}, 0 \leq x \leq 1\) and \(f(x)=\sqrt{x}, x \geq 1\), is (A) 1
View solution Problem 32
If \(f(x)=\int_{0}^{1} \frac{d t}{1+|x-t|}\), then \(f^{\prime}\left(\frac{1}{2}\right)\) is equal two (A) 1 (B) \(-1\) (C) \(\frac{1}{2}\) (D) 0
View solution Problem 33
The area bounded by the parabolas \(y^{2}=4 a(x+a)\) and \(y^{2}=-4 a(x-a)\) is (A) \(\frac{16}{3} a^{2}\) (B) \(\frac{8}{3} a^{2}\) (C) \(\frac{4}{3} a^{2}\) (
View solution