Definite Integration Question 294
Question: The area between the parabola $ y^{2}=4ax $ and $ x^{2}=8ay $ is
[RPET 1997]
Options:
A) $ \frac{8}{3}a^{2} $
B) $ \frac{4}{3}a^{2} $
C) $ \frac{32}{3}a^{2} $
D) $ \frac{16}{3}a^{2} $
Show Answer
Answer:
Correct Answer: C
Solution:
Required area $ A=\int_0^{a{2^{8/3}}} \sqrt{4ax},dx - \int_0^{a{2^{8/3}}} \frac{x^{2}}{8a},dx = \frac{32a^{2}}{3} $
BETA
