Volumes question (1 Viewer)

[Bank$]

This question is in A&A exercise 6.3 question 7

The base of a particular solid is x^2+y^2=4. Find the volume of the solid if every cross section perpendicular to the x axis is a parabolic segment with the axis of symmetry passing through the x axis and height the length of its base

I can do the question but u must use simpsons rule at one point to determine the area of the Parabolic segments. Personally i hate simpsons rule and wanted to integrate so I tried to get the equation of the parabola but it doesnt seem too come through, here is my working is in the right direction or totally of track lol

1. My equation for the binding parabola (ie the parabola that runs over the max height)
2. I sub in a value for x from the equation of the base but i dnt get a parabola
3. I equate the hight (2y) and try and integrate but it seems to be = to 0 as when u let x^2 - 4=U , the limits are equal :s

Thanks in Advance

justin

http://img267.imageshack.us/img267/4378/scansi3.jpg

 

[Trebla]

I'm not quite sure what you did, but you could have considered it this way:

Consider the general parabolic cross section in isolation. In other words consider the cross section on a completely separate set of axes. So put the parabola in the z-x axes for example such that it is symmetric over the z axis. You need to find the equation of the parabola, which is z = k(x - y)(x + y). Note that you're treating y as a constant. To find k, sub x = 0, z = 2y and you'll get k = - 2/y, hence the equation will be z = (- 2/y) (x - y)(x + y). You can now integrate in the form ∫z dx with limits -y to y and the area should work out in terms of y which we treat as a constant.

Try not to delve too much into the z axis here. You're making a simple problem much more complicated. The treatment of y being a constant in the integral over a three dimensional space is part of partial differential calculus - something you'll encounter in university mathematics.

 

[Bank$]

I'm not quite sure what you did, but you could have considered it this way:

Consider the general parabolic cross section in isolation. In other words consider the cross section on a completely separate set of axes. So put the parabola in the z-x axes for example such that it is symmetric over the z axis. You need to find the equation of the parabola, which is z = k(x - y)(x + y). Note that you're treating y as a constant. To find k, sub x = 0, z = 2y and you'll get k = - 2/y, hence the equation will be z = (- 2/y) (x - y)(x + y). You can now integrate in the form ∫z dx with limits -y to y and the area should work out in terms of y which we treat as a constant.

Thank u so much, yeah i saw my school teacher do that and thats what i was trying to mimic.

One question though c how we are making y a constant and having only 2 axis by doing that arent we then limiting to one parabola cause we are taking away the 3 dimensional aspect ?
What im trying to say is shouldnt the integral that we get no be able to be applied to all the other slices cause they only exist in 3 dimensions.

btw i did get the question right using that method its just I want to have a full understanding of the method.

Thanks again

Justin

 

[Trebla]

One question though c how we are making y a constant and having only 2 axis by doing that arent we then limiting to one parabola cause we are taking away the 3 dimensional aspect ?
What im trying to say is shouldnt the integral that we get no be able to be applied to all the other slices cause they only exist in 3 dimensions.

The point of obtaining the integral with the parabola was to find the general formula for the AREA of the cross section. We haven't found the VOLUME of the solid yet. Once you found your area, the volume can then be obtained by applying another round of integration.

The reason we treated y as a constant to obtain the area is because we are taking a "particular" y for a particular x according to the equation of the circle. Imagine the solid and a slice is taken from it. A certain slice will have a certain y value that determines its base length and height. We want to find the area of that particular slice, so we deal with it in isolation.
We treat y as a constant because the y is fixed for that particular slice. So when we go find the area, the area will be in terms of y, our "constant". But that only gives us the area of one particular slice.
Now to find the volume, y is no longer constant, so now we treat it as a variable because y varies in the 3 dimensional space of the solid. However we have eliminated the z for our area, so hence we only deal with x and y coordinates to sum the slices up for the volume integral.

Another way (probably a much easier way) to think of it is treating x as constant for a 'particular' slice (and from equation of a circle it follows that y is constant anyway, let the particular constants be x = x' and y = y' for that slice). If we completely ignore the x axis (since for any slice x is constant), we'll deal with the 2D z-y axes.
But remember x is a constant, hence y is a constant from the circle equation. So sub in the constant y' = root(4 - x'²) for the y values of the parabolic slice. Now integrate ∫z dy, where x' is a constant. Hence we'll obtain area in terms of x'. Now to find volume we have to generalise x and y as variables so treat it as a completely different problem as if you were given the area, and the integral for volume should follow.

If it's still confusing, try to think about it in terms of normal integration to find volume - we select a particular slice (or shell), derive a general formula for the cross-section area of each slice (or shell), hence we obtain the small change in volume dV and then sum up all the slices (or shells) as the thickness of each slice (or shell) approaches zero.

Note that for any equation in 2D space, we treat the 3rd dimension as constant. E.g. If we have an equation involving x-y-z such as x² + y² = z² if we let z be a constant such as 5, we'll get the familiar x-y 2D equation of
x² + y² = 25 (this is known as a level curve or a 'contour plot' if you remember your Geography lol). This is why if we treat one dimension as a constant then we can analyse the equation in a traditional 2D manner.

 

[Bank$]

The reason we treated y as a constant to obtain the area is because we are taking a "particular" y for a particular x according to the equation of the circle. Imagine the solid and a slice is taken from it. A certain slice will have a certain y value that determines its base length and height. We want to find the area of that particular slice, so we deal with it in isolation.
We treat y as a constant because the y is fixed for that particular slice. So when we go find the area, the area will be in terms of y, our "constant". But that only gives us the area of one particular slice.
Now to find the volume, y is no longer constant, so now we treat it as a variable because y varies in the 3 dimensional space of the solid. However we have eliminated the z for our area, so hence we only deal with x and y coordinates to sum the slices up for the volume integral.

Thanks alot that was a really gud explanation and i did another question the same way and got it right .

Justin