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

[tandoorichicken]

could someone please check my work on the following problem:

Find the volume under $z= 2x + y^2$ and above the region bounded by $x= y^2$ and $x= y^3$.

endpoints of region: (0,0) and (1,1)
$$V = \int_{0}^{1}\int_{0}^{1} (2x + y^2) \,dx\,dy = \int_{0}^{1} [x^2 + xy^2]|_{0}^{1}\,dy = (x^2 y + \frac{1}{3}xy^2)|_{(0,0)}^{(1,1)} = \frac{4}{3}$$

 

[dextercioby]

Shouldn't the points be in 3D...?And the integral be a triple one...?

Daniel.

 

[vincentchan]

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?

 

[dextercioby]

Shouldn't the volume of a domain D from R^{3} have the following form:
$$V_{D}=\iiint_{D} dx \ dy \ dz$$

Daniel.

 

[tandoorichicken]

you set your limits point incorrectly, they do not only depend on the end points

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

Somewhat confused.

 

[dextercioby]

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.

 

[dextercioby]

$$V = \int_{0}^{1}\int_{y^2}^{y^3} (2x + y^2) \,dx\,dy$$

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]

$$V = \int_{0}^{1}\int_{y^2}^{y^3} (2x + y^2) \,dx\,dy$$

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

$$V = \int_{0}^{1}\int_{y^3}^{y^2} (2x + y^2) \,dx\,dy$$

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

marlon

 

[tandoorichicken]

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

and got 4/3. I think this is right.

 

[dextercioby]

Nope,My Maple says it's wrong.It should come down to $$\frac{19}{210}$$

Daniel.

 

[Galileo]

$$\int_{0}^{1} [x^2 + xy^2]|_{y^3}^{y^2}\,dy = \int_{0}^{1} [(y^4 - y^4) - (y^6 - y^5)]\,dy$$

Check this step again.

 

[tandoorichicken]

OMG stupid arithmetic

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

Finally. Thanks guys.