Volume Using Cross-Sections

[JaiLit51853]

Looking for some help this question:
What is the volume of a solid whose base is bound between y=x^3 and the y-axis, if it has a cross section that is a rectangular prism whose height is double its base?

I know that y = x^3 and the y axis intersect at (0,0) and labelled the base as s, letting the height equal 2s.

But I am confused about how the question should be set up since the question doesn't specify which axis the cross sections are perpendicular to (not sure how I can tell) and the graph of y = x^3 goes on to positive infinity as x increases and negative infinity as x decreases (so the base doesn't seem to be completely bounded).

Any advice would be greatly appreciated!

 

[MarkFL]

Hello, and welcome to FMH!

I would agree with your observation that the base, as described, does not have a finite area. I would add a parameter, and say let the base also be bounded by the line \(y=a\) where \(0<a\). Our cross-sections would then presumably be perpendicular to the \(y\)-axis although we could also make them parallel, which would give us a different solid.

Making the cross-sections perpendicular to the \(y\)-axis, what would the base and height of an arbitrary cross-section be, in terms of \(y\)?

 

[JaiLit51853]

Hello, and welcome to FMH!

I would agree with your observation that the base, as described, does not have a finite area. I would add a parameter, and say let the base also be bounded by the line \(y=a\) where \(0<a\). Our cross-sections would then presumably be perpendicular to the \(y\)-axis although we could also make them parallel, which would give us a different solid.

Making the cross-sections perpendicular to the \(y\)-axis, what would the base and height of an arbitrary cross-section be, in terms of \(y\)?

Hi Mark, thanks for your response!

So then s=x, and the dimensions would be x by 2x, which I think turns into y^1/3 by 2y^1/3.
Since it's bound by y = a now, the integral would appear to be the integral from 0 to a of 2y^1/3 times y^1/3 dy.
That would equal the integral of from 0 to a of 2y^2/3,
Which equals 6/5 times y^5/3 evaluated from 0 to a.
That seems like it would just turn into 6/5 a^5/3.

Is that going in the right direction?

 

[JaiLit51853]

Oh, I should probably multiply this result by 2 since that 6/5 a^5/3 represents the volume above the x axis, so 12/5 a^5/3 would give the total volume of the solid.

 

[Dr.Peterson]

I presume you are also aware that "it has a cross section that is a rectangular prism" is nonsense as stated; you are taking it to mean "rectangle". A cross section is 2-dimensional, while a prism is 3-dimensional!

But the issues you brought up are more important, and I agree with MarkFL about what it most likely was supposed to mean. You'll have to state your assumptions, and if possible ask your teacher how to interpret the problem, which is quite poorly stated.

 

[MarkFL]

Yes, I took the statement about the rectangular prism as referring to an element of the total volume:

[MATH]dV=2x^2\,dy=2y^{\frac{2}{3}}\,dy[/MATH]
And so:

[MATH]V=2\int_0^a y^{\frac{2}{3}}\,dy=\frac{6}{5}a^{\frac{5}{3}}\quad\checkmark[/MATH]