Chapter 13: Q. 30 (page 1055)

Evaluate the iterated integral :
$\int_{0}^{1} \int_{0}^{3} \int_{0}^{\sin^{- 1} x} (x + y) d z d y d x$

Short Answer

Expert verified

$\frac{21 π}{8} - \frac{9}{2}$

Step by step solution

Step 1. Given information.

Given integral is :

$\int_{0}^{1} \int_{0}^{3} \int_{0}^{\sin^{- 1} x} (x + y) d z d y d x$

We have to evaluate it.

Step 2. Evaluate.

$\int_{0}^{1} \int_{0}^{3} \int_{0}^{\sin^{- 1} x} (x + y) d z d y d x$

$\begin{aligned} = \int_{0}^{1} \int_{0}^{3} [\int_{0}^{\sin^{- 1} x} (x + y) d z] d y d x \\ = \int_{0}^{1} \int_{0}^{3} [(x + y) \int_{0}^{\sin^{- 1} x} d z] d y d x \\ = \int_{0}^{1} \int_{0}^{3} [(x + y) [z]_{0}^{\sin^{- 1} x}] d y d x \\ = \int_{0}^{1} \int_{0}^{3} [(x + y) [\sin^{- 1} x]] d y d x \end{aligned}$

Step 3. Integrate with respect to y.

Integrate with respect to y

$\begin{aligned} = \int_{0}^{1} \int_{0}^{3} [(x + y) [\sin^{- 1} x]] d y d x \\ = \int_{0}^{1} [x \sin^{- 1} x \int_{0}^{3} d y + \sin^{- 1} x \int_{0}^{3} y d y] d x \\ = \int_{0}^{1} [x \sin^{- 1} x [y]_{0}^{3} + \sin^{- 1} x {[\frac{y^{2}}{2}]}_{0}^{3}] d x \\ = \int_{0}^{1} [x \sin^{- 1} x [3] + \sin^{- 1} x [\frac{3^{2}}{2}]] d x \\ = 3 \int_{0}^{1} x \sin^{- 1} x d x + \frac{9}{2} \int_{0}^{1} \sin^{- 1} x d x \end{aligned}$

Use integration by parts :

$\begin{aligned} = 3 [\frac{π}{8}] + \frac{9}{2} [\frac{π}{2} - 1] \\ = [\frac{3 π}{8} + \frac{9 π}{4} - \frac{9}{2}] \\ = \frac{21 π}{8} - \frac{9}{2} \end{aligned}$

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Evaluate the iterated integral :
$\int_{0}^{1} \int_{0}^{3} \int_{0}^{\sin^{- 1} x} (x + y) d z d y d x$

Short Answer

Step by step solution

Step 1. Given information.

Step 2. Evaluate.

Step 3. Integrate with respect to y.

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Evaluate the iterated integral :∫01 ∫03 ∫0sin−1⁡x (x+y)dzdydx

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Step by step solution

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Step 2. Evaluate.

Step 3. Integrate with respect to y.

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Evaluate the iterated integral :
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