Question

Evaluate ∫∫x^2+y^2 dxdy in the positive quadrant for which x+y<=1.

(a) \(\frac{1}{2} \)

(b) \(\frac{1}{3} \)

(c) \(\frac{1}{6} \)

(d) \(\frac{1}{12} \)

I had been asked this question by my school teacher while I was bunking the class.

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

Answer 1

Right choice is (c) \(\frac{1}{6} \)

The explanation is: In this

x: 0 to 1

y:0 to 1-x

\(\int_0^1 \int_0^{1-x} x^2+y^2 dxdy \)

\(= \int_0^1 x^2 y+ \frac{y^3}{3} dx \) (from 0 to 1-x)

\(= \int_0^1 x^2 (1-x)+ \frac{(1-x)^3}{3} dx \)

\(= \frac{1}{6}. \)