Definite Integration Question 24
Question: What is the area bounded by the curves $ x^{2}+y^{2}=9 $ and $ y^{2}=8x $ is
[DCE 1999]
Options:
A) 0
B) $ \frac{2\sqrt{2}}{3}+\frac{9\pi }{2}-9{{\sin }^{-1}}( \frac{1}{3} ) $
C) $ 16\pi $
D) None of these
Show Answer
Answer:
Correct Answer: B
Solution:
Solving the equations, $ x^{2}+y^{2}=9 $ …..(i)
$ y^{2}=8x $ …..(ii) Put $ y^{2}=8x $ in (i), $ x^{2}+8x-9=0 $
therefore $ (x+9)(x-1)=0 $ i.e., $ x=-9 $ or 1 $ x=-9 $ gives imaginary value of y for equation (ii) hence neglected.
$ \therefore A\equiv (1,0) $ and $ B\equiv (3,0) $
$ \therefore $ Required area = 2 times the hatched areas
$ =2[ \int_0^{1}{ydx}for\text{(ii)}+\int_1^{3}{ydxfor\text{(i)}} ] $
$ =2[ \int_0^{1}{2}\sqrt{2}{{(x)}^{1/2}}dx+\int_1^{3}{\sqrt{3^{2}-x^{2}}dx} ] $
$ =2[ 2\sqrt{2}\times ( \frac{{x^{3/2}}}{3/2} )_0^{1}+( \frac{x\sqrt{9-x^{2}}}{2}+\frac{9}{2}{{\sin }^{-1}}( \frac{x}{3} ) )_1^{3} ] $
$ =2[ \frac{4\sqrt{2}}{3}+\frac{9}{2}\times \frac{\pi }{2}-\frac{\sqrt{8}}{2}-\frac{9}{2}{{\sin }^{-1}}( \frac{1}{3} ) ] $
$ =\frac{2\sqrt{2}}{3}+\frac{9\pi }{2}-9{{\sin }^{-1}}( \frac{1}{3} ) $ .
BETA
