Question

Evaluate the value of \(\int\int_0^y \frac{2xy^5}{\sqrt{1+x^2 y^2-y^4}} dxdy\), where y varies from 0 to 1.

(a) ^11⁄12

(b) ^14⁄6

(c) ^11⁄6

(d) ^11⁄7

This question was posed to me in quiz.

My question is from Application of Double Integrals topic in division Multiple Integrals of Engineering Mathematics

Answer 1

Right option is (c) ^11⁄6

For explanation I would say: Given, f(x)=\(\int_0^{2a} \int_0^y \frac{2xy^5}{\sqrt{1+x^2 y^2-y^4}} dxdy\)

=\(\int_0^1 \int_0^y \frac{1}{y}  \frac{2xy^5}{\sqrt{(\frac{1-y^4}{y^2})+x^2)}} dxdy=\int_0^1 2y^4 |(\frac{1-y^4}{y^2})+x^2]|_0^y dy\)

=\(2\int_0^1 [y^3-\sqrt{1-y^4}y^3]dy=2\left [\frac{y^4}{4}-\frac{2}{3} (1-y^4)^{\frac{3}{2}}\right ] _1^0=2[\frac{1}{4}+\frac{2}{3}]=\frac{11}{6}\)