1-1-

Question Number 167588 by AlbertEinstein last updated on 20/Mar/22

1 + 1 = ¿

Answered by chhaythean last updated on 20/Mar/22

= \frac{1}{3} \int_{0}^{2 π} \int_{0}^{\frac{π}{4}} \sin φ \cos^{3} θ d φ d θ

= \frac{1}{3} \int_{0}^{2 π} \cos^{3} θ \times {[- \cos φ]}_{0}^{\frac{π}{4}} d θ

= \frac{1}{3} \times [- (\frac{\sqrt{2}}{2} - 1)] (\int_{0}^{2 π} \cos θ d θ - \int_{0}^{2 π} \sin^{2} θ d (\sin θ))

= \frac{1}{3} \times [- (\frac{\sqrt{2}}{2} - 1)] \times 0

So \begin{matrix} \int_{0}^{2 π} \int_{0}^{\frac{π}{4}} \int_{0}^{\cos θ} r^{2} \sin φ drd φ d θ = 0 \end{matrix}