Question

Evaluate \(\int_0^1\int_0^{\sqrt{1-x^2}}\int_0^{\sqrt{1-x^2-y^2}}xyzdxdydz.\)

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

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

(c) \(\frac{1}{24} \)

(d) \(\frac{1}{48} \)

I have been asked this question in class test.

My query is from Volume Integrals in portion Vector Integral Calculus of Engineering Mathematics

Answer 1

Correct choice is (c) \(\frac{1}{24} \)

The best explanation: \(\int_0^1 \int_0^{\sqrt{1-x^2}}\int_0^{\sqrt{1-x^2-y^2}}xyzdxdydz=  \int_0^1 \int_0^{\sqrt{1-x^2}} \frac{xyz^2}{2} dxdy \) from 0 to \( \sqrt{1-x^2-y^2} \)

\(= \frac{1}{2} \int_0^1 \frac{xy^2}{2} – \frac{x}{2} – \frac{xy^4}{4} dx \) from 0 to \(\sqrt{1-x^2} \)

\(= \frac{1}{2} \int_0^1 \frac{(x-x^3)}{2}-\frac{(x^3-x^5)}{2}-\frac{(x+x^5-2x^3)}{4} dx\)

\(= \frac{1}{4}\big\{\frac{1}{6} – \frac{1}{2} + \frac{1}{2}\big\} \)

\(= \frac{1}{24}. \)