Definite Integration Question 188
Question: The area of the region R={$(x,y):| x |\le | y | $ and $ x^{2}+y^{2}\le 1$} is
Options:
A) $ \frac{3\pi }{8} $ sq. unit
B) $ \frac{5\pi }{8} $ sq. unit
C) $ \frac{\pi }{2} $ sq. unit
D) $ \frac{\pi }{8} $ sq. unit
Show Answer
Answer:
Correct Answer: C
Solution:
[c] required area = area of the shaded region = 4 (area of the shaded region in first quadrant) $ =4\int_0^{1/\sqrt{2}}{(y_1-y_2)dx=4\int_0^{1/\sqrt{2}}{(\sqrt{1-x^{2}}-x)dx}} $
$ =4[ \frac{1}{2}\times \sqrt{1-x^{2}}+\frac{1}{2}{{\sin }^{-1}}x-\frac{x^{2}}{2} ]_0^{1/\sqrt{2}}=\frac{\pi }{2} $ sq. unit
BETA
