Double Integration: Volume Under z = 2x + y^2

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Homework Help Overview

The discussion revolves around finding the volume under the surface defined by z = 2x + y², above the region bounded by the curves x = y² and x = y³. The endpoints of the region are specified as (0,0) and (1,1).

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants explore the setup of the double integral and question the limits of integration based on the bounding curves. Some express confusion about the correct order of integration and the interpretation of the curves in relation to the volume calculation.

Discussion Status

There are multiple interpretations of the limits of integration, with some participants suggesting corrections to the original setup. Guidance has been offered regarding the integration order and the relationship between the curves, but no consensus has been reached on the correct volume calculation.

Contextual Notes

Participants are discussing the implications of using double versus triple integrals for volume calculations in three-dimensional space. There is also mention of potential arithmetic errors in the calculations presented.

tandoorichicken
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could someone please check my work on the following problem:

Find the volume under [itex]z= 2x + y^2[/itex] and above the region bounded by [itex]x= y^2[/itex] and [itex]x= y^3[/itex].

endpoints of region: (0,0) and (1,1)
[tex]V = \int_{0}^{1}\int_{0}^{1} (2x + y^2) \,dx\,dy = \int_{0}^{1} [x^2 + xy^2]|_{0}^{1}\,dy = (x^2 y + \frac{1}{3}xy^2)|_{(0,0)}^{(1,1)} = \frac{4}{3}[/tex]
 
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Shouldn't the points be in 3D...?And the integral be a triple one...?

Daniel.
 
you set your limits point incorrectly, they do not only depend on the end points, (0,0) and (1,1), but also depend on the curve x=y^2 and x=y^3

And the integral be a triple one...?[/tex]
can you explain your thought?
 
Shouldn't the volume of a domain D from R^{3} have the following form:
[tex]V_{D}=\iiint_{D} dx \ dy \ dz[/tex]

Daniel.
 
Last edited:
you set your limits point incorrectly, they do not only depend on the end points

So it should be:
[tex]V = \int_{0}^{1}\int_{y^3}^{y^2} (2x + y^2) \,dx\,dy = \int_{0}^{1} [x^2 + xy^2]|_{y^3}^{y^2}\,dy = \int_{0}^{1} [(y^4 - y^4) - (y^6 - y^5)]\,dy = \int_{0}^{1} y^5 - y^6 \,dy = [\frac{1}{6}y^6 - \frac{1}{7}y^7]|_{0}^{1} = \fract{1}{6} - \frac{1}{7} = \frac{1}{42} ?[/tex]

Somewhat confused.
 
Are u sure that the values of the square (y²) are less than the values of the cube (y³) ...?If so,then the first integral is correct.Unfortunately,the 3-rd integral is incorrect,and hence everything that follows.

Daniel.

P.S.Pay attention to the signs.
 
marlon said:
[tex]V = \int_{0}^{1}\int_{y^2}^{y^3} (2x + y^2) \,dx\,dy[/tex]

This is what it should be. Keep in mind that you first integrate to x and then to y. This is because x is given as a function of y and NOT the other way around. In your figure, you need to look at the Y-axis as the HORIZONTAL axis (rotation of 90°). Then you will see that for y going from 0 to 1, the y^3 is ABOVE the y^2
regards
marlon


What...? What's your definition for "ABOVE"...?

To the OP:disregard Marlon's post.Your integration (the one you started with was good,the arithmetics that followed was sloppy).

Daniel.
 
marlon said:
[tex]V = \int_{0}^{1}\int_{y^2}^{y^3} (2x + y^2) \,dx\,dy[/tex]

This is what it should be. Keep in mind that you first integrate to x and then to y. This is because x is given as a function of y and NOT the other way around. In your figure, you need to look at the Y-axis as the HORIZONTAL axis (rotation of 90°). Then you will see that for y going from 0 to 1, the y^3 is ABOVE the y^2

regards
marlon

[tex]V = \int_{0}^{1}\int_{y^3}^{y^2} (2x + y^2) \,dx\,dy[/tex]

this is what it should be ... I was wrong. Sorry bout that. I mixed the two functions

marlon
 
okay i fixed the math in
[tex]V = \int_{0}^{1}\int_{y^3}^{y^2} (2x + y^2) \,dx\,dy[/tex]

and got 4/3. I think this is right.
 
  • #10
Nope,My Maple says it's wrong.It should come down to [tex]\frac{19}{210}[/tex]

Daniel.
 
  • #11
tandoorichicken said:
[tex]\int_{0}^{1} [x^2 + xy^2]|_{y^3}^{y^2}\,dy = \int_{0}^{1} [(y^4 - y^4) - (y^6 - y^5)]\,dy[/tex]
Check this step again.
 
  • #12
OMG stupid arithmetic

[tex]\int_{0}^{1} [x^2 + xy^2]|_{y^3}^{y^2}\,dy = \int_{0}^{1} [(y^4 + y^4) - (y^6 + y^5)]\,dy = (\frac{2}{5}y^5 - \frac{1}{7}y^7 - \frac{1}{6}y^6)|_{0}^{1} = \frac{19}{210}[/tex]

Finally. Thanks guys.
 

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