Exercise 8.1 Solutions
Chapter 8: Application of Integrals
Q1
Find the area of the region bounded by the ellipse
\[ \frac{x^2}{16} + \frac{y^2}{9} = 1 \]
Step-by-Step Solution
1. Identify the Ellipse:
The equation is \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\).
Here, \(a^2 = 16 \Rightarrow a = 4\) and \(b^2 = 9 \Rightarrow b = 3\).
The equation is \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\).
Here, \(a^2 = 16 \Rightarrow a = 4\) and \(b^2 = 9 \Rightarrow b = 3\).
2. Exploit Symmetry:
The ellipse is symmetrical about both axes. The total area is 4 times the area in the first quadrant. \[ \text{Total Area} = 4 \times \int_0^a y \, dx \] From the equation: \(\frac{y^2}{9} = 1 – \frac{x^2}{16} \Rightarrow y = \frac{3}{4}\sqrt{16 – x^2}\) (in 1st quadrant).
The ellipse is symmetrical about both axes. The total area is 4 times the area in the first quadrant. \[ \text{Total Area} = 4 \times \int_0^a y \, dx \] From the equation: \(\frac{y^2}{9} = 1 – \frac{x^2}{16} \Rightarrow y = \frac{3}{4}\sqrt{16 – x^2}\) (in 1st quadrant).
3. Integrate:
\[ A = 4 \int_0^4 \frac{3}{4}\sqrt{4^2 – x^2} \, dx = 3 \int_0^4 \sqrt{4^2 – x^2} \, dx \] Using formula \(\int \sqrt{a^2-x^2} dx = \frac{x}{2}\sqrt{a^2-x^2} + \frac{a^2}{2}\sin^{-1}\frac{x}{a}\): \[ A = 3 \left[ \frac{x}{2}\sqrt{16-x^2} + \frac{16}{2}\sin^{-1}\frac{x}{4} \right]_0^4 \] \[ A = 3 \left[ (0 + 8\sin^{-1}(1)) – (0 + 0) \right] \] \[ A = 3 \left[ 8 \times \frac{\pi}{2} \right] = 3 \times 4\pi = 12\pi \]
\[ A = 4 \int_0^4 \frac{3}{4}\sqrt{4^2 – x^2} \, dx = 3 \int_0^4 \sqrt{4^2 – x^2} \, dx \] Using formula \(\int \sqrt{a^2-x^2} dx = \frac{x}{2}\sqrt{a^2-x^2} + \frac{a^2}{2}\sin^{-1}\frac{x}{a}\): \[ A = 3 \left[ \frac{x}{2}\sqrt{16-x^2} + \frac{16}{2}\sin^{-1}\frac{x}{4} \right]_0^4 \] \[ A = 3 \left[ (0 + 8\sin^{-1}(1)) – (0 + 0) \right] \] \[ A = 3 \left[ 8 \times \frac{\pi}{2} \right] = 3 \times 4\pi = 12\pi \]
Total Area = \(12\pi\) sq. units
Q2
Find the area of the region bounded by the ellipse
\[ \frac{x^2}{4} + \frac{y^2}{9} = 1 \]
Step-by-Step Solution
1. Identify parameters:
\(a^2 = 4 \Rightarrow a = 2\), \(b^2 = 9 \Rightarrow b = 3\). Note that here \(b > a\) (vertical ellipse), but the integration logic remains the same.
\(a^2 = 4 \Rightarrow a = 2\), \(b^2 = 9 \Rightarrow b = 3\). Note that here \(b > a\) (vertical ellipse), but the integration logic remains the same.
2. Setup Integral:
Total Area \( = 4 \int_0^2 y \, dx \).
\(y = \frac{3}{2}\sqrt{4 – x^2}\).
Total Area \( = 4 \int_0^2 y \, dx \).
\(y = \frac{3}{2}\sqrt{4 – x^2}\).
3. Calculate:
\[ A = 4 \int_0^2 \frac{3}{2}\sqrt{2^2 – x^2} \, dx = 6 \int_0^2 \sqrt{2^2 – x^2} \, dx \] \[ A = 6 \left[ \frac{x}{2}\sqrt{4-x^2} + \frac{4}{2}\sin^{-1}\frac{x}{2} \right]_0^2 \] \[ A = 6 \left[ (0 + 2\sin^{-1}(1)) – 0 \right] \] \[ A = 6 \left[ 2 \times \frac{\pi}{2} \right] = 6\pi \]
\[ A = 4 \int_0^2 \frac{3}{2}\sqrt{2^2 – x^2} \, dx = 6 \int_0^2 \sqrt{2^2 – x^2} \, dx \] \[ A = 6 \left[ \frac{x}{2}\sqrt{4-x^2} + \frac{4}{2}\sin^{-1}\frac{x}{2} \right]_0^2 \] \[ A = 6 \left[ (0 + 2\sin^{-1}(1)) – 0 \right] \] \[ A = 6 \left[ 2 \times \frac{\pi}{2} \right] = 6\pi \]
Total Area = \(6\pi\) sq. units
Q3
MCQ: Area in 1st quadrant bounded by circle \(x^2 + y^2 = 4\), \(x=0, x=2\)
Choose the correct answer:
(A) \(\pi\) (B) \(\frac{\pi}{2}\) (C) \(\frac{\pi}{3}\) (D) \(\frac{\pi}{4}\)
(A) \(\pi\) (B) \(\frac{\pi}{2}\) (C) \(\frac{\pi}{3}\) (D) \(\frac{\pi}{4}\)
Solution
Analysis:
The curve \(x^2+y^2=4\) is a circle with radius \(r=2\).
The limits \(x=0\) to \(x=2\) cover exactly the first quadrant of this circle.
The curve \(x^2+y^2=4\) is a circle with radius \(r=2\).
The limits \(x=0\) to \(x=2\) cover exactly the first quadrant of this circle.
Calculation:
\[ A = \int_0^2 y \, dx = \int_0^2 \sqrt{4-x^2} \, dx \] \[ A = \left[ \frac{x}{2}\sqrt{4-x^2} + \frac{4}{2}\sin^{-1}\frac{x}{2} \right]_0^2 \] \[ A = \left[ 2\sin^{-1}(1) – 0 \right] = 2(\frac{\pi}{2}) = \pi \]
\[ A = \int_0^2 y \, dx = \int_0^2 \sqrt{4-x^2} \, dx \] \[ A = \left[ \frac{x}{2}\sqrt{4-x^2} + \frac{4}{2}\sin^{-1}\frac{x}{2} \right]_0^2 \] \[ A = \left[ 2\sin^{-1}(1) – 0 \right] = 2(\frac{\pi}{2}) = \pi \]
Correct Option: (A) \(\pi\)
Q4
MCQ: Area bounded by curve \(y^2 = 4x\), y-axis and line \(y=3\)
Choose the correct answer:
(A) 2 (B) \(\frac{9}{4}\) (C) \(\frac{9}{3}\) (D) \(\frac{9}{2}\)
(A) 2 (B) \(\frac{9}{4}\) (C) \(\frac{9}{3}\) (D) \(\frac{9}{2}\)
Solution
Analysis:
The curve is a parabola opening to the right. The region is bounded by the y-axis (\(x=0\)) and \(y=3\). It is easier to integrate with respect to \(y\).
The curve is a parabola opening to the right. The region is bounded by the y-axis (\(x=0\)) and \(y=3\). It is easier to integrate with respect to \(y\).
Setup Integral w.r.t y:
Given \(y^2 = 4x \Rightarrow x = \frac{y^2}{4}\).
Limits are from \(y=0\) to \(y=3\). \[ A = \int_0^3 x \, dy = \int_0^3 \frac{y^2}{4} \, dy \]
Given \(y^2 = 4x \Rightarrow x = \frac{y^2}{4}\).
Limits are from \(y=0\) to \(y=3\). \[ A = \int_0^3 x \, dy = \int_0^3 \frac{y^2}{4} \, dy \]
Calculation:
\[ A = \frac{1}{4} \left[ \frac{y^3}{3} \right]_0^3 \] \[ A = \frac{1}{12} [ 3^3 – 0 ] = \frac{27}{12} \] Simplifying \(\frac{27}{12}\) by dividing by 3 gives \(\frac{9}{4}\).
\[ A = \frac{1}{4} \left[ \frac{y^3}{3} \right]_0^3 \] \[ A = \frac{1}{12} [ 3^3 – 0 ] = \frac{27}{12} \] Simplifying \(\frac{27}{12}\) by dividing by 3 gives \(\frac{9}{4}\).
Correct Option: (B) \(\frac{9}{4}\)