Volume of a square by integration

[mares09]

Use integration tecniques to find the volume of a square extruded along a parabola of the form y= 6 - (x^2) from y=6 to y=0.

Im stuck here because i dont know if i should consider the side of the square to be y= 6 - (x^2) or y= (6 - (x^2)) / 2.

Any help would be greatly appreciated.

 

[galactus]

Use neither. I don't know what you mean by extruded, but I assume you mean the cross sections along the y-axis are squares. To 'extrude' means
to push through, like dough through a pasta maker. I don't know how that applies to this problem.


You are integrating along the y-axis. Therefore, \(\displaystyle \L\\x=\sqrt{6-y}\) is a side of the square.

That will be one side of the parabola. Multiply by two.

\(\displaystyle \L\\2\int_{0}^{6}(\sqrt{6-y})^{2}dy\)

 

[mares09]

Thank you!! yeah I am a little confused about what is asked on this problem. Im not sure is worded correctly. I appreciate all your help today.

Thank you.