Definite Integration Question 314
Question: Area included between the two curves $ y^{2}=4ax $ and $ x^{2}=4ay, $ is
[SCRA 1986; Roorkee 1984; RPET 1999; Kerala (Engg.) 2002, 05]
Options:
A) $ \frac{32}{3}a^{2} $ sq. unit
B) $ \frac{16}{3} $ sq. unit
C) $ \frac{32}{3} $ sq. unit
D) $ \frac{16}{3}a^{2} $ sq. unit
Show Answer
Answer:
Correct Answer: D
Solution:
Solving the two equations, we have $ x^{4}=64a^{3}x $
$ \Rightarrow x=0,4a $ Required area = $ \int_0^{4a}{2{a^{1/2}}{x^{1/2}}dx-\int_0^{4a}{\frac{x^{2}}{4a}dx}} $
$ =\frac{32}{3}a^{2}-\frac{16}{3}a^{2}=\frac{16}{3}a^{2} $ sq. unit.
BETA
