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Area using double integals

  • Thread starter oceansoft
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  • #1
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Hello everyone am new to this forum my name is lucy and i hope i can help and get help from other i got this tutorial question that i seem to keep getting wrong :s i hope someone can help me :) the question is

Use a double integral to calculate the area of the region in the positive quadrant of the xy-plane bounded above by y=1, to the left by y= 1 -x and to the right by y=(1-x)^3

first i reaggare y=1-x and y=(1-x)^3 to get

x=1-y and x=y1/3 + 1

to get the double integal that looks like this

[tex]\oint^{1}_{0}[/tex] [tex]\oint^{y^1/3 + 1}_{1-y}[/tex] 1 dx dy

= [tex]\oint^{1}_{0}[/tex] y^1/3 + 1 - 1 + y
= [3/4 y^4/3 + 1/2 y^2] between 1 and 0
= 4/9 + 1/2 - 0
= 17/18

Sorry for the mess am still trying to get a hang of this, am been at this question for ages am pretty bad at those but if any one could give a little bit of help it will be of a lot of help to me thanks alot :)
 
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Answers and Replies

  • #2
699
6
Hello everyone am new to this forum my name is lucy and i hope i can help and get help from other i got this tutorial question that i seem to keep getting wrong :s i hope someone can help me

... bounded above by y=1, to the left by y= 1 -x and to the right by y=(1-x)^3 ...
Hi Lucy,

It seems to me that there is a mistake in the wording of this question. If you draw a plot of your boundaries the y=1-x line is to the right of the y=(1-x)^3 curve, not to the left as written. As written, the area is zero, but I doubt this is the correct question. The question would make more sense if the curve were y=(x-1)^3.
 
  • #3
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Hi Lucy,

It seems to me that there is a mistake in the wording of this question. If you draw a plot of your boundaries the y=1-x line is to the right of the y=(1-x)^3 curve, not to the left as written. As written, the area is zero, but I doubt this is the correct question. The question would make more sense if the curve were y=(x-1)^3.
Hey thanks for the help but i just miss type that when I was working it out on paper i did use y=(x-1)^3 and that what it is in the question but sill got the wrong answer. do u know what else i might of done wrong?
 
  • #4
699
6
Hey thanks for the help but i just miss type that when I was working it out on paper i did use y=(x-1)^3 and that what it is in the question but sill got the wrong answer. do u know what else i might of done wrong?
Oh, OK. For that problem, you did it correctly, but just made a slight error at the end.

I get the following copying from your work.

= [3/4 y^4/3 + 1/2 y^2] between 1 and 0
= 3/4 + 1/2 - 0
= 5/4
 
  • #5
11
0
Oh, OK. For that problem, you did it correctly, but just made a slight error at the end.

I get the following copying from your work.

= [3/4 y^4/3 + 1/2 y^2] between 1 and 0
= 3/4 + 1/2 - 0
= 5/4
Yeh thanks i know where i went wrong when i type 1^4/3 in i didn't put the bracket around 4/3. btw are u such i havn't gone wrong anywhere else if not thanks for the help
 
  • #6
699
6
Yeh thanks i know where i went wrong when i type 1^4/3 in i didn't put the bracket around 4/3. btw are u such i havn't gone wrong anywhere else if not thanks for the help
As far as i can tell you did it correctly. The final number also makes physical sense. Just looking at the plot, it's clear that the area should be just a little greater than one.
 

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