Evaluate the iterated integral: a)int_0^3 int_0^(pi/3) int_0^4 r xx cos theta drd theta dz b)int_0^(pi/2) int_0^3 int_0^(4 - z) zdrdzd theta

rocedwrp 2021-09-02 Answered

Evaluate the iterated integral:
a) $\int_{0}^{3} \int_{0}^{\frac{π}{3}} \int_{0}^{4} r \times \cos θ d r d θ d z$

b) $\int_{0}^{\frac{π}{2}} \int_{0}^{3} \int_{0}^{4 - z} z d r d z d θ$

c) $\int_{0}^{\frac{π}{2}} \int_{0}^{\frac{π}{2}} \int_{0}^{2} ρ^{2} d ρ d θ d ϕ$

d) $\int_{0}^{\frac{π}{4}} \int_{0}^{\frac{π}{4}} \int_{0}^{\cos (ϕ)} \cos θ d ρ d ϕ d θ$

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Expert Answer

2abehn

Answered 2021-09-03 Author has 88 answers

a) $\int_{0}^{3} \int_{0}^{\frac{π}{3}} \int_{0}^{4} r \times \cos θ d r d θ d z =$

$= \int_{0}^{3} \int_{0}^{\frac{π}{3}} \cos θ {[γ^{\frac{2}{2}}]}_{0}^{4} d θ d z =$

$= \int_{0}^{3} \int_{0}^{\frac{π}{3}} 8 \cos θ \times d θ \times d z =$

$= \int_{0}^{3} 8 \times {[\sin θ]}_{0}^{\frac{π}{3}} d z =$

$= \int_{0}^{3} 8 \times \frac{\sqrt{3}}{2} d z =$

$= 4 \sqrt{3} \int_{0}^{3} d z =$

$= 4 \sqrt{3} {[z]}_{0}^{3} =$

$= 12 \sqrt{3}$

b) $\int_{0}^{\frac{π}{2}} \int_{0}^{3} \int_{0}^{4 - z} z d r d z d θ =$

$= \int_{0}^{\frac{π}{2}} \int_{0}^{3} z \times {[γ]}_{0}^{4 - z} d z \times d θ =$

$= \int_{0}^{\frac{π}{2}} \int_{0}^{3} z \times [(4 - 2) - 0] d z d θ =$

$= \int_{0}^{\frac{π}{2}} \int_{0}^{3} 4 z - z^{2} \times d z d θ =$

$= \int_{0}^{\frac{π}{2}} {[(4 \frac{z^{2}}{2}) - \frac{z^{3}}{3}]}_{0}^{3} d θ =$

$= \int_{0}^{\frac{π}{2}} (18 - 9) \times d θ =$

$= 9 \int_{0}^{\frac{π}{2}} d θ =$

$= 9 {[θ]}_{0}^{\frac{π}{2}} =$

$= \frac{9 π}{2}$

c)

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