Question

Evaluate ∫∫xy dxdy over the region bounded by x axis, ordinate x=2a and the curve x^2=4ay.

(a) \(\frac{a^4}{3} \)

(b) \(\frac{a^4}{6} \)

(c) \(\frac{a^3}{3} \)

(d) \(\frac{a^2}{6} \)

This question was addressed to me in a job interview.

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

Answer 1

Correct answer is (a) \(\frac{a^4}{3} \)

To elaborate: Both the curves meet at (2a,a).

Therefore,

\(x:0 \rightarrow 2a \)

\(y: 0 \rightarrow \frac{x^2}{4a} \)

\(\int_0^2 a \int_0^{\frac{x^2}{4}}a xydxdy \)

\(= \int_0^2 a  \frac{xy^2}{2} dx \) (from 0 to \( \frac{x^2}{4a}) \)

\( = \frac{1}{2} \int_0^{2a} \frac{x^5}{(16a^2 )} dx \)

\( = \frac{a^4}{3}. \)