Question

Evaluate  ∫xy dxdy over the positive quadrant of the circle x^2+y^2=a^2.

(a) \(\frac{a^4}{8} \)

(b) \(\frac{a^4}{4} \)

(c) \(\frac{a^2}{8} \)

(d) \(\frac{a^2}{4} \)

I got this question during an online interview.

The question is from Surface Integrals topic in section Vector Integral Calculus of Engineering Mathematics

Answer 1

Correct option is (a) \(\frac{a^4}{8} \)

To elaborate: In the positive quadrant of the circle,

\(y: 0 \rightarrow a \)

\(x: o \rightarrow \sqrt{a^2-x^2} \)

Therefore the integral is

\(\displaystyle\int_0^a \int_0^{\sqrt{a^2-x^2}}xydxdy \)

\( = \int_0^a \frac{yx^2}{2} dy \) (from 0 to \(\sqrt{a^2-x^2}) \)

\( = \frac{1}{2} \int_0^a y(a^2-y^2)  dy= \frac{a^4}{8}. \)