Answered: "Evaluate the following integrals in cylindrical coordinates. ∫04∫022∫x1−xe−x2−y2dydxdz"

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Answered question

2021-10-09

Evaluate the following integrals in cylindrical coordinates.

\int_{0}^{4} \int_{0}^{\frac{\sqrt{2}}{2}} \int_{x}^{\sqrt{1 - x}} e^{- x^{2} - y^{2}} d y d x d z

Alix Ortiz

Skilled2021-10-10Added 109 answers

We rewrite the integral in cylindrical coordinates. We know that

r^{2} = x^{2} + y^{2}, 0 \leq r \leq 1, 0 \leq z \leq 4, 0 \leq θ \leq \frac{π}{4}

. Therefore,

\int_{0}^{\frac{π}{4}} \int_{0}^{1} \int_{0}^{4} (e^{- r^{2}}) r d z d r d θ = \int_{0}^{\frac{π}{4}} \int_{0}^{1} r e^{- r^{2}} {[z]}_{0}^{4} d r d θ

= \int_{0}^{\frac{π}{4}} \int_{0}^{1} 4 r e^{- r^{2}} d r d θ

= \int_{0}^{\frac{π}{4}} (4 \int_{0}^{- 1} - \frac{e^{u}}{2} d u) d θ

= \int_{0}^{\frac{π}{4}} (2 \int_{- 1}^{0} e^{u} d u) d θ

= \int_{0}^{\frac{π}{4}} (2 {[e^{u}]}_{- 1}^{0}) d θ

= \int_{0}^{\frac{π}{4}} 2 (1 - \frac{1}{e}) d θ

= 2 (1 - \frac{1}{e}) {[θ]}_{0}^{\frac{π}{4}}

= \frac{π}{2} (1 - \frac{1}{e})

Result:

\frac{π}{2} (1 - \frac{1}{e})

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