MHB Volumn of parabola and line with perpendicular cross sections being a square.

[karush]

The base of a solid is the region bounded by the parabola
$$y^2=4x$$, and the line $$x=2$$ .

Each plane section perpendicular to the x-axis is square.
(I assume this means the cross-section of the solid will be square)
then we are not revolving but slicing.

The volume of the solid is? (the ans is 32) I looked at an example the book but didn't understand how the integral was set up.

 

[MarkFL]

I would begin by switching variables simply because I find it easier to work the problem that way. So, the base is bounded by:

$$y=\frac{x^2}{4}$$

and

$$y=2$$

Draw you a sketch of the region. Now, we are slicing perpendicularly to the $y$-axis. So, what will the sides $s$ of the square be, first in terms of $x$, and from that, in terms of $y$?. Then, having found that, the volume of an arbitrary slice is:

$$dV=s^2\,dy$$

Can you proceed?

 

[karush]

assume the change variables makes this a function in terms of $x$

and $s$ then would be just $2x$ or the area of $4x^2$

but isn't this going to $\int_a^b{A(y)}dy$

 

[MarkFL]

Yes, on both counts. You have correctly found that the side length of each square if $2x$, and so we have:

$$dV=(2x)^2\,dy=4x^2\,dy$$

Now, is there any way we can relate $x^2$ to $y$?

 

[karush]

Yes, on both counts. You have correctly found that the side length of each square if $2x$, and so we have:

$$dV=(2x)^2\,dy=4x^2\,dy$$

Now, is there any way we can relate $x^2$ to $y$?

well if we can use $y=\frac{x^2}{4}$ then $16y=4x^2$

so $\displaystyle\int_0^2{16y} \text{ dy} =32$

this looks too easy..(Wondering)

 

[MarkFL]

well if we can use $y=\frac{x^2}{4}$ then $16y=4x^2$

so $\displaystyle\int_0^2{16y} \text{ dy} =32$

this looks too easy..(Wondering)

That's all there is to it. :D

With these types of problems, the hard part is usually not the calculus, but setting up the integral. (Thinking)