Question

Find the area inside a ellipse of minor-radius ‘b’ and major-radius ‘a’.

(a) –^4⁄3 a^2

(b) –^4⁄3 ab^2

(c) ^4⁄3 ab

(d) –^4⁄3

I had been asked this question in an interview for internship.

I want to ask this question from Application of Double Integrals topic in chapter Multiple Integrals of Engineering Mathematics

Answer 1

The correct answer is (c) ^4⁄3 ab

To elaborate: Equation of ellipse is given by \(\frac{x^2}{a^2} +\frac{y^2}{b^2}=1\)

\(\int_{-b}^b \int_{-a}^a [\frac{x^2}{a^2}+\frac{y^2}{b^2}-1]dxdy=\int_{-b}^b [\frac{x^3}{3a^2}+\frac{xy^2}{b^2}-x]_a^{-a} dy\)

\(\int_{-b}^b [\frac{a}{3}+\frac{ay^2}{b^2} -a+\frac{a}{3}+\frac{ay^2}{b^2}-a] dy\)

=\(\int_{-b}^b [\frac{2a}{3}+\frac{2ay^2}{b^2}-2a]  dy=[\frac{2ay}{3}+\frac{2ay^3}{3b^2}-2ay] _b^{-b}\)

=\(\frac{4ab}{3}+\frac{4ab}{3}-4ab=-\frac{4}{3} ab\)