Answered: V2/2 (VT- /1-x2 et-°dy dx dz

Mathematics For Machine Technology

8th Edition

ISBN:9781337798310

Author:Peterson, John.

Publisher:Peterson, John.

Chapter58: Achievement Review—section Five

Section: Chapter Questions

Problem 30AR: Determine dimension x to 3 decimal places.

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Question

Integrals in cylindrical coordinates Evaluate the following integral in cylindrical coordinates. The figures, if given, illustrate the region of integration.

Expert Solution

Step 1

The integral is given as,

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

The limits of the integral is,

$0 \leq x \leq \frac{\sqrt{2}}{2}$ , $x \leq y \leq \sqrt{1 - x^{2}}$ , $0 \leq z \leq 4 .$

The equation of the circle is $x^{2} + y^{2} = 1$ which implies $0 \leq r \leq 1, \frac{π}{4} \leq θ \leq \frac{π}{2}$ .

$\begin{array}{rcl} f (x, y, z) & = & \int_{\frac{π}{4}}^{\frac{π}{2}} \int_{0}^{1} \int_{0}^{4} e^{- r^{2}} r d z d r d θ \\ = & \int_{\frac{π}{4}}^{\frac{π}{2}} \int_{0}^{1} 4 r e^{- r^{2}} d r d θ . \end{array}$

Step 2

Substitute $r^{2} = u$ .

$2 r d r = d u .$

Change the limits ,

$r = 0 \to u = 0 r = 1 \to u = 1$

The integral becomes,

$\begin{array}{rcl} f (x, y, z) & = & \int_{\frac{π}{4}}^{\frac{π}{2}} [\int_{0}^{1} 4 r e^{- r^{2}} d r] d θ . \\ = & \int_{\frac{π}{4}}^{\frac{π}{2}} [2 \int_{0}^{1} e^{- u} d u] d θ . \end{array}$

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