Homework Help: Vof solid with circular base&right isoceloes perpendicular cross sections

1. Oct 12, 2008

Alixx69

[[URGENT]]Vof solid with circular base&right isoceloes perpendicular cross sections

1. The problem statement, all variables and given/known data

so there is a circle described by the equation x^2+y^2=1 with right isosceles triangles perpendicular to it. each of the triangles is touching the circle with one of its LEGS, with the 90 degree angle and one of the smaller angles touching the perimeter of the circle. now use an integral to find the volume of the solid produced by the adding all of the areas of the infinitely thin triangles from y = -1 to 1.

(Just in case my interpretation wasn't as clear to you as it is to me, the actual question is:
"The base of the solid is the disk x^2+y^2 <= 1. The cross-sections by planes perpendicular to the y axis between y= -1 and y=1 are isosceles right triangles with one LEG on the disk")

2. Relevant equations

V=int from a to b(A(x)dx)

3. The attempt at a solution

so i got,
A(x)=int from -1 to 1(height of the triangle*base of the triangle)
base = x

and this is where i get tripped up,
if the base, being one of the LEGS of the triangle is x, shouldnt the height(the other LEG of the isosceles triangle) be x too?

V(x)=int from -1 to 1((.5x^2)dx)...
but that would mean that , using .5[(x^3)/3] as the antiderivitive and evaluating from -1 to 1, the answer is 1/3

but if you think about it in a different way , its a half of a can that has a r of 1 and a h of 2
and using (pi*r^2*h)/2 you would get 3.14 which is pi

where am i going wrong with this?

Last edited: Oct 12, 2008
2. Oct 13, 2008

Dick

3. Oct 13, 2008

Staff: Mentor

Re: [[URGENT]]Vof solid with circular base&right isoceloes perpendicular cross sectio

If you multiply the base of the triangle by its height, you'll get the area of a square, not a triangle. (I'm using the fact, which you recognize, that each triangle is isosceles, making its two legs the same length.)

What you need to do is to get an expression that gives you the volume of one of the thin triangular pieces, and to do this, you need to understand all of the dimensions of it. The volume of one of the triangular pieces is its area times its thickness. This expression will give you the integrand, the expression in the integral. The thickness plays a role in helping you determine the limits of integration, which I don't think you understand. You wrote them as -1 to 1. Which variable = -1? Which variable = 1?

You're partly right here. Since the triangles are isosceles, the height of each triangle will be equal to its base. But the length of the base is not x. The base of each triangle extends from a point (x1, y) on the left side of the circle to a point (x2, y) on the right side. Can you find an expression that represents this length? When you find it, you can find the area of a typical triangle, which you can multiply by the thickness to get the volume of a typical volume element.
I get a value that's quite a bit bigger.
That seems like a reasonable approach, but doesn't apply to this problem. If you sliced the can in half diagonally, the sliced edge would be an ellipse that touched the circular base at just one point. In this problem, every triangle's hypotenuse touches the circular base.

4. Oct 13, 2008

Alixx69

Re: [[URGENT]]Vof solid with circular base&right isoceloes perpendicular cross sectio

i meant the hight being being height and the base being the area of the circle, but what you said got me thinking...
so the height being the value y of the solved equation x^2+y^2=r^2(r is the radius) which is y=sqrt(r^2-x^2)
and the base being the area of the circle which is pi*r^2
then it would be int from -1 to 1((pi*r^2)(sqrt(r^2-x^2))dx?

no the circles LEG touches the base,, from the side, one of the cross sections looks like:
|\
|...\
|......\
******
with the lines made of "|" and "*" being the same length
and the line of "\" being the hypotenuse and of a longer length
(the periods were used for spacing reasons)

5. Oct 13, 2008

Staff: Mentor

Re: [[URGENT]]Vof solid with circular base&right isoceloes perpendicular cross sectio

The height of the triangle is not the y value on the circle. The height of each triangle is the same as the base. BTW, you know what the radius of the circle is, so there's no point in introducing another variable r when you already know it.

I've given you some pointers on how to set up the integral in my previous post, so take a careful look at what I wrote.

I think you meant the triangle's leg touches the base. I have a similar picture, but it's 3D. In any case, the solid described in this problem is not half of a cylinder.

6. Oct 13, 2008

Alixx69

Re: [[URGENT]]Vof solid with circular base&right isoceloes perpendicular cross sectio

For the triangle cross sections
V = A*h
h = dx
A=.*l*w
l = w = sqrt(1-x^2)
is that right?
would that make the integral
V=int from -1 to 1(.5(2*sqrt(1-x^2)))?
that gives me an answer of .5pi

but im so sure the answer is pi because vv
how can it not be half a cylinder, if you cut a cylinder in half, diagonally, you have two identical pieces with the volume of Pi*r^2 all over 2

and because

in my book it says the answer can be found by
V = int from 0 to 2(2Xsqrt(2*r*x-x^2))dx
and the answer to that is pi
but i have no idea how they got there, how did they?
and why are they using 0 to 2 instead of -1 to 1?

7. Oct 13, 2008

Staff: Mentor

Re: [[URGENT]]Vof solid with circular base&right isoceloes perpendicular cross sectio

No, the thickness is not dx. The bases of the triangles are all parallel to the x-axis, so the thickness of each one is $$\Delta$$y.

Your formula for area is wrong. If you meant .5*l*w, that would be correct. You have a period in what you wrote, so maybe you meant to write .5 but got in a hurry and left it out.

No. The base of each triangle extends from the left side of the circle to its right side, so w = x - (-x) = 2x.
The height of each triangle is the same as its width, so h = 2x.
The area of each triangle is 1/2 * (2x) * (2x) = 2x^2.
The thickness of each triangle is $$\Delta$$y.
So $$\Delta$$ = 2x^2 * $$\Delta$$y.
Since your volume element involves $$\Delta$$y, you will need to convert 2x^2 into an expression in y. This should be easy to do, since the endpoints of each base are on the circle, whose equation you know.
Because vv?
I've given you a reason why the solid in this problem is not a half cylinder, but here it is again. If you cut a cylinder in half diagonally, the edge of the cut is in the shape of an ellipse. This ellipse touches the base at a single point, (1, 0, 0). The ellipse touches a single point at the top of the cylinder (-1, 0, 2).

For the solid you're working with, every triangle touches the right half of the circular base (assuming the right angles of the triangles are on the left, which is the way you and I have represented things). This is different from what it would be if you sliced the cylinder in half diagonally.
That's a good question. Integrating from 0 to 2 makes no sense in this problem. Two possibilities are 1) the answer you're looking at isn't the one for the problem you're working on, or 2) the answer in your book is wrong, which I guarantee has happened in a lot of books.
Mark

8. Oct 13, 2008

Staff: Mentor

Re: [[URGENT]]Vof solid with circular base&right isoceloes perpendicular cross sectio

No, the thickness is not dx. The bases of the triangles are all parallel to the x-axis, so the thickness of each one is delta y.

Your formula for area is wrong. If you meant .5*l*w, that would be correct. You have a period in what you wrote, so maybe you meant to write .5 but got in a hurry and left it out.

No. The base of each triangle extends from the left side of the circle to its right side, so w = x - (-x) = 2x.
The height of each triangle is the same as its width, so h = 2x.
The area of each triangle is 1/2 * (2x) * (2x) = 2x^2.
The thickness of each triangle is delta y.
So delta V = 2x^2 * delta y.
Since your volume element involves delta y, you will need to convert 2x^2 into an expression in y. This should be easy to do, since the endpoints of each base are on the circle, whose equation you know.

Because vv?
I've given you a reason why the solid in this problem is not a half cylinder, but here it is again. If you cut a cylinder in half diagonally, the edge of the cut is in the shape of an ellipse. This ellipse touches the base at a single point, (1, 0, 0). The ellipse touches a single point at the top of the cylinder (-1, 0, 2).

For the solid you're working with, every triangle touches the right half of the circular base (assuming the right angles of the triangles are on the left, which is the way you and I have represented things). This is different from what it would be if you sliced the cylinder in half diagonally.
That's a good question. Integrating from 0 to 2 makes no sense in this problem. Two possibilities are 1) the answer you're looking at isn't the one for the problem you're working on, or 2) the answer in your book is wrong, which I guarantee has happened in a lot of books.
Mark