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Q. 46

Expert-verified

Calculus

Found in: Page 1004

Book edition 1st

Author(s) Peter Kohn, Laura Taalman

Pages 1155 pages

ISBN 9781429241861

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Short Answer

Evaluate each of the double integrals in Exercises 37–54 as iterated integrals.
$\int \int_{R} x^{3} e^{x^{2} y} d A, w h e r e R = {(x, y) | 0 \leq x \leq 4 a n d - 1 \leq y \leq 1}$

The value is $\frac{e^{32} - 2 e^{16} + 1}{2 e^{16}}$

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1. Given Information:

Given double integrals :

$\int \int_{R} x^{3} e^{x^{2} y} d A, w h e r e R = {(x, y) | 0 \leq x \leq 4 a n d - 1 \leq y \leq 1}$

We want to evaluate each of the double integrals as iterated integrals.

Step 2. Solution:

Using Fubini's Theorem:

$\int \int_{R} x^{3} e^{x^{2} y} d A = \int_{0}^{4} \int_{- 1}^{1} x^{3} e^{x^{2} y} d y d x E v a l u a t i o n p r o c e d u r e f o r t h e i t e r a t e d i n t e g r a l w e g e t = \int_{0}^{4} (\int_{- 1}^{1} x^{3} e^{x^{2} y} d y) d x = \int_{0}^{4} x^{3} (\int_{- 1}^{1} e^{x^{2} y} d y) d x$

$F i r s t w e s o l v e \int_{- 1}^{1} e^{x^{2} y} d y P u t x^{2} y = t s o w e h a v e d y = \frac{d t}{x^{2}} W h e n y = 1 t h e n t = x^{2} W h e n y = - 1 t h e n t = - x^{2} S o i n t e g r a l b e c o m e s : = \int_{- x^{2}}^{x^{2}} \frac{e^{t}}{x^{2}} d t = \frac{1}{x^{2}} \int_{- x^{2}}^{x^{2}} e^{t} d t = \frac{1}{x^{2}} {[e^{t}]}_{- x^{2}}^{x^{2}} = \frac{e^{x^{2}} - e^{- x^{2}}}{x^{2}} N o w d o u b l e i n t e g r a l b e c o m e s : = \int_{0}^{4} x^{3} (\frac{e^{x^{2}} - e^{- x^{2}}}{x^{2}}) d x = \int_{0}^{4} x (e^{x^{2}} - e^{- x^{2}}) d x = \int_{0}^{4} x e^{x^{2}} d x - \int_{0}^{4} x e^{- x^{2}} d x = I_{1} - I_{2}$

$S o l v e I_{1} w e g e t \int_{0}^{4} x e^{x^{2}} d x P u t x^{2} = t s o w e h a v e x d x = \frac{d t}{2} w h e n x = 4 t h e n t = 16 w h e n x = 0 t h e n t = 0 I_{1} b e c o m e s \int_{0}^{16} \frac{e^{t}}{2} d t = {[\frac{e^{t}}{2}]}_{0}^{16} = \frac{e^{16} - 1}{2} S o l v e I_{2} w e g e t P u t - x^{2} = t s o w e h a v e x d x = - \frac{d t}{2} w h e n x = 4 t h e n t = - 16 w h e n x = 1 t h e n t = 0 I_{1} b e c o m e s \int_{0}^{- 16} - \frac{e^{t}}{2} d t = {[- \frac{e^{t}}{2}]}_{0}^{- 16} = - \frac{e^{- 16} - 1}{2} = \frac{1 - e^{- 16}}{2} S o w e h a v e I_{1} - I_{2} = \frac{e^{16} - 1}{2} - \frac{1 - e^{- 16}}{2} = \frac{e^{16} - 1}{2} - \frac{e^{16} - 1}{2 e^{16}} = \frac{e^{32} - 2 e^{16} + 1}{2 e^{16}}$

Most popular questions for Math Textbooks

Let $f (x, y, z)$ be a continuous function of three variables, let localid="1650352548375" $Ω_{y z} = {(y, z) | a \leq y \leq b a n d h_{1} (y) \leq z \leq h_{2} (y)}$ be a set of points in the $y z$ -plane, and let localid="1650354983053" $Ω = {(x, y, z) | (y, z) \in Ω_{y z} a n d g_{1} (y, z) \leq x \leq g_{2} (y, z)}$ be a set of points in $3$ -space. Find an iterated triple integral equal to the triple integral localid="1650353288865" $\underset{Ω}{∭} f (x, y, z) d V$ . How would your answer change if localid="1650352747263" $Ω_{y z} = {(y, z) | a \leq z \leq b a n d h_{1} (z) \leq x \leq h_{2} (z)}$ ?

Evaluate each of the double integrals in Exercises 37–54 as iterated integrals.

$\int \int_{R} x \sin x \cos y d A, w h e r e R = {(x, y) | - 3 \leq x \leq 2 a n d - 2 \leq y \leq 2}$

What is the difference between a double integral and an iterated integral?

Find the volume between the graph of the given function and the xy-plane over the specified rectangle in the xy-plane

$f (x, y) = - 2 x^{2} y^{3} W h e r e R = {(x, y) | - 2 \leq x \leq 3, - 1 \leq y \leq 5}$

Evaluate each of the double integral in the exercise 37-54 as iterated integrals

$\int \int_{R} (2 - 3 x^{2} + y^{3}) d A W h e r e R = {(x, y) | - 3 \leq x \leq 2, 3 \leq y \leq 5}$

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