Finding the volume of a solid using cross sections

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fbellman

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Hi everybody! I was doing great in this chapter up until we started doing volume.. agh. Anyway, here's the problem I'm struggling with:

The base of a solid is bounded by y=x^3, y=0, and x=1. Find the volume of the solid for each of the following cross sections (taken perpendicular to the y-axis): (a) squares, (b) semi-circles, (c) equilateral triangles, and (d) semi-ellipses whose heights are twice the lengths of their bases.

So:
1) What the heck IS a semiellipse??? Alright, I figured this out! :)
2) To solve these problems, I know that you're supposed to take the antiderivative of the area of the shape. And since you're rotating around the y-axis I figured you have to put y=x^3 in the form of x=f(y). So that's x= cuberoot y. For the square, I thought you would integrate cuberoot y on the interval -1 to 1. But this didn't work out. The answer's 1/10, if anyone knows how to get there!
3) As for the other shapes......

Thanks in advance for any help!!!
 
fbellman said:
1) What the heck IS a semiellipse?
Click to expand...
A "semi-circle" is half of a circle. I'd guess a semi-ellipse to be half of an ellipse.

Eliz.
 
Well, I gathered as much! I looked it up elsewhere on the web and I see what an ellipse is now. Thanks!

Can anyone help me out with #2??? I know that you have to take the antiderivative of the area, so for squares you would take the antiderivative of .. side^2. But this doesn't seem to work (aka I get the wrong answer). And

area for equilateral triangles is: (side)^2 * (sqrt 3)/4

Help???
 
It is phrased strangely but for part a I would have interpreted part as when you look at a slice when you are standing in quadrant IV looking north you see a square with a side starting from the line, ending at x=1. How you can have a square, a semicircle or an equilateral triangle twice as high as it is wide I don't know. Unless That applies only to the semi ellipse. Maybe someone else has a different interpretation.
Part a) would start with a
square 1 high by 1 wide with a volume = 1 times dy
After you have removed squares up to y=.125 the square you see is 1/2 by 1/2 square. You would be integrating
(1-x)^2 dy
 
Hi Gene! Thanks a lot for helping me out.

I spent a lot of time trying to figure out the whole "twice as high as it is wide" thing too. Like you said, it's not really possible to have a square, semicircle or eq. triangle that fits the description, so that part must apply just to the semiellipse, I think. I hope!

For the square, I know that I'm integrating the area of one square. Area of a square is side squared, so I thought I'd be taking the integral of the area but... that didn't work out. The answer for the square one is 1/10, but I have NO idea how to get there.

Any ideas??
 
All the bases are 1-x so you are integrating
(1-x)^2dy
y=x^3 so dy=3x^2dx then you are integrating
(1-x)^2*(3x^2*dx) from x=0 to 1
That should indeed come out as .1 cubic whatevers.

The others are the same except for the area formula. Picture yourself standing looking up at the shape, moving it up the y axis and replace (1-x)^2 with the area formula for whatever the shape is. The area A is fixed for any point on the y axis and the incremental volume is dV=A*dy=A*3x^2*dx
 
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