Definite Integration Question 422
Question: The area of the loop of the curve $ ay^{2}=x^{2}(a-x) $ is
Options:
A) $ 4a^{2} $ sq. units
B) $ \frac{8a^{2}}{15} $ sq. units
C) $ \frac{16a^{2}}{9} $ sq. units
D) None of these
Show Answer
Answer:
Correct Answer: B
Solution:
[b] $ ay^{2}=x^{2}(a-x) $ or $ y=\pm x\sqrt{\frac{a-x}{a}} $ Curve tracing: $ y=x\sqrt{\frac{a-x}{a}} $ We must have $ x\le a $ For $ 0<x\le a,y>0 $ and $ x<0,y<0 $ Also $ y=0\Rightarrow x=0,a $ Curve is symmetrical about x-axis. When $ x\to -\infty ,y\to -\infty $ Also, it can be verified that y has only point of maxima for $ 0<x<a. $ Area $ =2\int\limits_0^{a}{x\sqrt{\frac{a-x}{a}}dx\sqrt{\frac{a-x}{a}}=t} $
$ \Rightarrow $ $ 1-\frac{x}{a}=t^{2} $ or $ x=a(1-t^{2}) $
$ \Rightarrow A=2\int\limits_1^{0}{a(1-t^{2})t(-2at)dt} $
$ =4a^{2}\int\limits_0^{1}{(t^{2}-t^{4})dt} $
$ =4a^{2}{{[ \frac{t^{3}}{3}=\frac{t^{5}}{5} ]}^{1}}_0 $
$ =4a^{2}[ \frac{1}{3}-\frac{1}{5} ] $
$ =\frac{8a^{2}}{15} $ sq. units.
BETA
