Help-me-0-0-3cos-sin-d-d-

Question Number 179344 by neinhaltsieger369 last updated on 28/Oct/22

Help - me!

\int_{0}^{π} \int_{0}^{3 \cos ϕ} θ \sin ϕ d θ d ϕ

Commented by CElcedricjunior last updated on 29/Oct/22

k = \int_{0}^{π} \int_{0}^{3 c o s \emptyset} θ s i n \emptyset d θ d \emptyset

k = \int_{0}^{π} s i n \emptyset {[\frac{1}{2} θ^{2}]}_{0}^{3 s i n \emptyset} d \emptyset

k = \frac{9}{2} \int_{0}^{π} si \overset{3}{n \emptyset d \emptyset} = \frac{9}{2} \int_{0}^{π} (s i n \emptyset - s i n \emptyset c o \overset{2}{s \emptyset}) d \emptyset

k = \frac{9}{2} {[- \cos \emptyset + \frac{co \overset{3}{s \emptyset}}{3}]}_{0}^{π} = \frac{9}{2} (1 - \frac{1}{3})

k = 3

d^{'} o \overset{`}{u}

\int_{0}^{π} \int_{0}^{3 s i n \emptyset} θ \sin \emptyset d θ d \emptyset = 3 u m

\dots \dots \dots \dots . . l e c e l e b r e c e d r i c j u n i o r \dots \dots . .

Commented by neinhaltsieger369 last updated on 01/Nov/22

Gracias

Answered by mr W last updated on 28/Oct/22

= \int_{0}^{π} \sin ϕ (\int_{0}^{3 \cos ϕ} θ d θ) d ϕ

= \int_{0}^{π} \sin ϕ (\frac{9 \cos^{2} ϕ}{2}) d ϕ

= - \frac{9}{2} \int_{0}^{π} \cos^{2} ϕ d (\cos ϕ)

= - \frac{9}{2} {[\frac{\cos^{3} ϕ}{3}]}_{0}^{π}

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

= 3

Commented by neinhaltsieger369 last updated on 29/Oct/22

Thank you!