• :00Days
  • :00Hours
  • :00Mins
  • 00Seconds
A new era for learning is coming soonSign up for free
Log In Start studying!

Select your language

Suggested languages for you:
Answers without the blur. Sign up and see all textbooks for free! Illustration

Q. 45

Found in: Page 1004


Book edition 1st
Author(s) Peter Kohn, Laura Taalman
Pages 1155 pages
ISBN 9781429241861

Answers without the blur.

Just sign up for free and you're in.


Short Answer

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

Rxx+ydA,whereR={(x,y) | 1x4 and 0y3}

The value is -8ln(2)+92+72ln(7)

See the step by step solution

Step by Step Solution

Step 1. Given Information:

Given double integrals :

Rxx+ydA,whereR={(x,y) | 1x4 and 0y3}

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

Step 2. Solution:

Using Fubini's Theorem

Rxx+ydA=1403xx+ydy dxEvaluation procedure for the iterated integral we get=1403xx+ydy dx=14x031x+ydy dx

First we solve:

031x+ydy=ln(x+t)03=ln(x+3)-ln xSo integral becomes:=14xln(x+3)-ln xdx=14xln(x+3)dx-14xln xdx=I1-I2Solve I1 we havePut x+3=t so we get dx=dtwhen x=4 then t=7when x=1 then t=4I1 becomes 47(t-3) ln(t) dtApply integration By parts:=ln(t)t22-3t-12t-6dt47=ln(t)t22-3t-12t22-6t47=8ln(2)+34+72ln(7)Solve I2 we have14xln xdxApply integration By parts:=ln(x)x22-x2dx14=ln(x)x22-x2414=16 ln(2)-154So we haveI1-I2=8ln(2)+34+72ln(7)-16 ln(2)+154=-8ln(2)+92+72ln(7)

Recommended explanations on Math Textbooks

94% of StudySmarter users get better grades.

Sign up for free
94% of StudySmarter users get better grades.