Chapter 13: Q. 28 (page 1055)

Evaluate the iterated integral :
$\int_{0}^{1} \int_{0}^{3 - 3 y} \int_{0}^{6 y} \frac{x}{y} d z d x d y$

Short Answer

Expert verified

$9$

Step by step solution

Step 1. Given information.

Given integral is :

$\int_{0}^{1} \int_{0}^{3 - 3 y} \int_{0}^{6 y} \frac{x}{y} d z d x d y$

We have to evaluate it.

Step 2. Evaluate.

Given $\int_{0}^{1} \int_{0}^{3 - 3 y} \int_{0}^{6 y} \frac{x}{y} d z d x d y$

$\begin{aligned} = \int_{0}^{1} \int_{0}^{3 - 3 y} [(\int_{0}^{6 y} \frac{x}{y} d z) d x d y] \\ = \int_{0}^{1} \int_{0}^{3 - 3 y} [(\frac{x}{y} \int_{0}^{6 y} d z) d x d y] \\ = \int_{0}^{1} \int_{0}^{3 - 3 y} [\frac{x}{y} (z)_{0}^{6 y}] d x d y \\ = \int_{0}^{1} \int_{0}^{3 - 3 y} [\frac{x}{y} (6 y)] d x d y \\ = \int_{0}^{1} \int_{0}^{3 - 3 y} (6 x) d x d y \end{aligned}$

Step 3. Integrate with respect to z.

Integrate with respect to z

$= \int_{0}^{1} [(\int_{0}^{3 - 3 y} 6 x d x) d y] = \int_{0}^{1} (6 {(\frac{x^{2}}{2})}_{0}^{3 - 3 y}) d y = \int_{0}^{1} {(3 x^{2})}_{0}^{3 - 3 y} d y \begin{aligned} = \int_{0}^{1} 3 (3 - 3 y)^{2} d y \\ = \int_{0}^{1} 3 (9 - 18 y + 9 y^{2}) d y \\ = 27 \int_{0}^{1} d y - 54 \int_{0}^{1} y d y + 27 \int_{0}^{1} y^{2} d y \\ = 27 (y)_{0}^{1} - 54 {(\frac{y^{2}}{2})}_{0}^{1} + 27 {(\frac{y^{3}}{3})}_{0}^{1} \\ = 27 - 27 + 9 \\ = 9 \end{aligned}$

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Evaluate the iterated integral :
$\int_{0}^{1} \int_{0}^{3 - 3 y} \int_{0}^{6 y} \frac{x}{y} d z d x d y$

Short Answer

Step by step solution

Step 1. Given information.

Step 2. Evaluate.

Step 3. Integrate with respect to z.

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Evaluate the iterated integral :∫01 ∫03−3y ∫06y xydzdxdy

Short Answer

Step by step solution

Step 1. Given information.

Step 2. Evaluate.

Step 3. Integrate with respect to z.

One App. One Place for Learning.

Most popular questions from this chapter

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Evaluate the iterated integral :
$\int_{0}^{1} \int_{0}^{3 - 3 y} \int_{0}^{6 y} \frac{x}{y} d z d x d y$