volume by slicing

[Jaina]

All I have to do in this problem is find a formua for the area of the cross-sections of the solid perpendicular to the x-axis. The equation of the circle is x² + y² = 1. The solid lies between planes perpendicular to the x-axis at x=-1 and x=1. In each case, the cross-sections perpendicular to the x-axis between these planes run from y= - √(1-x²) to y=√(1-x²). There's a picture, but I don't know how to post that.

a) The cross-sections are squares with bases in the xy-plane.

b) The cross-sections are equilateral triangles with bases in the xy-plane.

For some reason, I got the first one with the circle, but those squares and triangles are messing with my head.

 

[Unco]

G'day, Jaina.

I'm certainly confused by your description (my fault). "The cross-sections perpendicular to the x-axis between these planes run from y= - √(1-x²) to y=√(1-x²)" gets me as these aren't ordinates but are the bottom and top halves of the circle, respectively, so I need some help there...

Does it look anything like this?

             y /|\       
       |        |        |
             *  +  *       x^2 + y^2 = 1
       |  *     |     *  |
        *       |       *  
       *        |        *
   <---+--------+--------+------->
       -1       |        1     x 
       |*       |       *|
          *     |     *   
       |     *  +  *     |  
                |        
       |        |        | x=1
       x=-1

Which part are we rotating?

Thanks.

 

[Jaina]

Hello!

I wish it was my description, then I might understand it. Sadly, it's straight out of the book. My book has problems with words, and making them understandable. :-(

Let me try to put it in my own words and see if I make any sense. I have a basic graph of a circle, x² + y² = 1. I'm finding the volume from -1 to 1 on the x-axis.

The book doesn't say a thing about rotating the graph at all... it wants me just to find a formula for the area of the slice they tell me to use. We're not finding the actual volume in this problem, it's just doing the first step.

So for the first part, they want a slice that's a square that kind of sits on top of the plane (the xy plane is laying flat, so to speak). It looks like somebody set a picture of a circle on the ground, then took a piece of paper and tried to stand it up on the circle.

I think that I need to figure out the formula for the area of the square and relate that somehow to the equation of a circle. At least, that's the impression I'm getting. Does my description make a little more sense? I think I just confused myself even more...

 

[soroban]

Hello, Jaina!

Find a formua for the area of the cross-sections of the solid perpendicular to the x-axis.
The equation of the circle is $\displaystyle x^2\,+\,y^2\,=\,1$
The solid lies between planes perpendicular to the x-axis at $\displaystyle x=-1$ and $\displaystyle x=1$.

In each case, the cross-sections perpendicular to the x-axis between these planes
run from y= - √(1-x²) to y=√(1-x²). . We don't need this information.

a) The cross-sections are squares with bases in the xy-plane.

b) The cross-sections are equilateral triangles with bases in the xy-plane.

                |
               ***  P
           *    |   |*      The cross-section is taken at PQ.
          *     |  y| *
         *      |   |  *         The length of PQ is 2y.
      - -*- - - + - + -*- -
         *      |   |  *
          *     |  y| *
           *    |   |*
               ***  Q
                |

Think of a loaf of bread with a circular base,
The slices are made perpendicular to the x-axis (like PQ).


In part (a), the slices are all squares with side $\displaystyle PQ = 2y$.

The 'face' of the slice has area: $\displaystyle (2y)^2\,=\,4y^2$.
The 'thickness' of the slice is $\displaystyle dx$.
Hence, the volume of one slice is: .$\displaystyle dV\:=\:4y^2\,dx$

Since $\displaystyle y\:=\:\sqrt{1\,-\,x^2}$, we have: .$\displaystyle dV\:=\:4(\sqrt{1\,-\,x^2})^2\,dx\:=\;4(1-x^2)\,dx$


To find the total volume, we just "add up the slices":

. . \(\displaystyle \L V\;=\;4\int^{\;\;\;1}_{-1}(1 - x^2)\,dx\)

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

The procedure is the same for part (b).

Since the cross-sections are equilateral triangles,
. . we have a different formula.

            *
           /|\
          / | \
         /  |h \
        /   |   \
       /    |    \
      *-----+-----* 
         y     y

The base is still $\displaystyle 2y$.
We find that the height is $\displaystyle \sqrt{3}y$.

The area of the triangle is: .$\displaystyle A\:=\:\frac{1}{2}(2y)(\sqrt{3}y)\:=\;\sqrt{3}y^2$

The volume of the triangular slice is: .$\displaystyle dV\:=\:\sqrt{3}y^2\,dx\:=\:\sqrt{3}(1\,-\,x^2)\,dx$

The total volume is: .\(\displaystyle \L\sqrt{3}\int^{\;\;\;1}_{-1}(1\,-\,x^2)\,dx\)