Evaluate the iterated integral: a) ∫03∫0π3∫04r×cos⁡θdrdθdz b) ∫0π2∫03∫04−zzdrdzdθ c) ∫0π2∫0π2∫02ρ2dρdθdϕ d) ∫0π4∫0π4∫0cos⁡(ϕ)cos⁡θdρdϕdθ

a) ∫03∫0π3∫04r×cos⁡θdrdθdz= =∫03∫0π3cos⁡θ[γ22]04dθdz= =∫03∫0π38cos⁡θ×dθ×dz= =∫038×[sin⁡θ]0π3dz= =∫038×32dz= =43∫03dz= =43[z]03= =123 b) ∫0π2∫03∫04−zzdrdzdθ= =∫0π2∫03z×[γ]04−zdz×dθ= =∫0π2∫03z×[(4–2)–0]dzdθ= =∫0π2∫034z–z2×dzdθ= =∫0π2[(4z22)–z33]03dθ= =∫0π2(18−9)×dθ= =9∫0π2dθ= =9[θ]0π2= =9π2 c)

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rocedwrp

Answered question

2021-09-02

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

Answer & Explanation

2abehn

Skilled2021-09-03Added 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) $\int_{0}^{\frac{π}{2}} \int_{0}^{3} \int_{0}^{4 - z} z d r d z d θ =$

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

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

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

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

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

$= 9 \int_{0}^{\frac{π}{2}} d θ =$

$= 9 {[θ]}_{0}^{\frac{π}{2}} =$

$= \frac{9 π}{2}$

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