Inter 2nd Year Maths Exercise 8a Solutions

Practicing the AP Inter 2nd Year Maths Study Material and AP Inter 2nd Year Maths Chapter 8 Application of Integrals Solutions Exercise 8a will help students to clear their doubts quickly.

Inter 2nd Year Maths Application of Integrals Solutions Exercise 8a

Application of Integrals Exercise 8a Solutions

Question 1.
Find the area of the region bounded by the ellipse \(\frac{x^2}{16}+\frac{y^2}{9}=1\)
Solution:
Given ellipse \(\frac{x^2}{16}+\frac{y^2}{9}=1\) (a > b)
Comparing to \(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\)
a² = 16
⇒ a = 4
b² = 9
⇒ b = 3
From ellipse is \(\frac{x^2}{16}+\frac{y^2}{9}=1\)
⇒ \(\frac{y^2}{9}=1-\frac{x^2}{16}\)
⇒ \(\frac{y^2}{9}=\frac{16-x^2}{16}\)
⇒ \(y^2=\frac{9}{16}\left(16-x^2\right)\)
⇒ y = \(\sqrt{\frac{9}{16}\left(16-x^2\right)}\)
⇒ y = \(\frac{3}{4} \sqrt{16-x^2}\)

The area of OAB = \(\int_0^4 y d x\)
= \(\int_0^4 \frac{3}{4} \sqrt{16-x^2} d x\) [∵ \(\int \sqrt{a^2-x^2} d x=\frac{x}{2} \sqrt{a^2-x^2}+\frac{a^2}{2} \sin ^{-1} \frac{x}{a}\)]
= \(\frac{3}{4}\left[\begin{array}{l}
\frac{4}{2} \sqrt{4^2-4^2}+\frac{16}{2} \sin ^{-1}\left(\frac{4}{4}\right) \\
-\left(\frac{0}{2} \sqrt{4^2-0^2}+\frac{16}{2} \sin ^{-1}\left(\frac{0}{4}\right)\right)
\end{array}\right]\)
= \(\frac{3}{4} \times 8 \times \frac{\pi}{2}\)
= 3π
The total area of the ellipse = 4 × OAB
= 4 × 3π
= 12π

Question 2.
Find the area of the region bounded by the ellipse \(\frac{x^2}{4}+\frac{y^2}{9}=1\)
Solution:
Given ellipse is \(\frac{x^2}{4}+\frac{y^2}{9}=1\) (a < b)
Comparing to \(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\)
a² = 4
⇒ a = 2
b² = 9
⇒ b = 3
From ellipse \(\frac{x^2}{4}+\frac{y^2}{9}=1\)
⇒ \(\frac{y^2}{9}=1-\frac{x^2}{4}\)
⇒ \(\frac{y^2}{9}=\frac{4-x^2}{4}\)
⇒ \(y^2=\frac{9}{4}\left(4-x^2\right)\)
⇒ y = \(\frac{3}{2} \sqrt{4-x^2}\)

Area of OAB w.r.to X-axis
Area of OAB = \(\int_0^2 y d x\)
= \(\int_0^2 \frac{3}{2} \sqrt{4-x^2} d x\)
= \(\frac{3}{2}\left[\frac{x}{2} \sqrt{2^2-x^2}+\frac{2^2}{2} \sin ^{-1} \frac{x}{2}\right]_0^2\) [∵ \(\int \sqrt{a^2-x^2} d x=\frac{x}{2} \sqrt{a^2-x^2}+\frac{a^2}{2} \sin ^{-1} \frac{x}{a}\)]
= \(\frac{3}{2}\left[\begin{array}{l}
\left(\frac{2}{2} \sqrt{2^2-2^2}+\frac{4}{2} \sin ^{-1}\left(\frac{2}{2}\right)\right) \\
-\left(\frac{0}{2} \sqrt{2^2-0^2}+\frac{2^2}{2} \sin ^{-1}\left(\frac{0}{2}\right)\right)
\end{array}\right]\)
= \(\frac{3}{2} \times 2 \times \frac{\pi}{2}\)
= \(\frac{3 \pi}{2}\)
Total area of ellipse = 4 × \(\frac{3 \pi}{2}\) = 6π