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Q. 46

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.

Rx3ex2ydA,where R={(x,y)| 0x4 and 1y1}

The value is e32-2e16+12e16

See the step by step solution

Step by Step Solution

Step 1. Given Information:

Given double integrals :

Rx3ex2ydA,where R={(x,y)| 0x4 and 1y1}

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

Step 2. Solution:

Using Fubini's Theorem:

Rx3ex2ydA=04-11x3ex2y dy dxEvaluation procedure for the iterated integral we get=04-11x3ex2y dy dx=04x3-11ex2y dy dx

First we solve-11ex2y dyPut x2y=t so we have dy=dtx2When y=1 then t=x2When y=-1 then t=-x2So integral becomes:=-x2x2etx2 dt=1x2-x2x2et dt=1x2et-x2x2=ex2-e-x2x2Now double integral becomes:=04x3ex2-e-x2x2dx=04xex2-e-x2dx=04xex2 dx-04xe-x2dx=I1-I2

Solve I1we get04xex2 dxPut x2=t so we have x dx=dt2when x=4 then t=16when x=0 then t=0I1 becomes 016et2 dt=et2016=e16-12Solve I2 we getPut -x2=t so we have x dx=-dt2when x=4 then t=-16when x=1 then t=0I1 becomes 0-16-et2 dt=-et20-16=-e-16-12=1-e-162So we haveI1-I2=e16-12-1-e-162=e16-12-e16-12e16=e32-2e16+12e16

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