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metacogitans

Volume Integrals?

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For finding the volume between a multivariable line integral (x,y) and a single variable integral along the x axis, does this look right?

 

( Volume 0,0x+h,yf(x+h,y) 0,0x+h,0f(x+h) - Volume 0,0x,yf(x,y) 0,0x,0f(x) ) / h = x,0x,y f(x,y)

 

In particular I'm looking at f(x,y) parabaloids in the format z=axn+bym if that makes a difference; you can assume that f(x,y)-f(y)=f(x)

 

I don't know if that's right or not, I was thinking of the volume as consisting of infinitesimal integral prisms, and following that the volume of a prism is its base multiplied by its height, the difference in volume from one of the infinitesimal integral prisms divided by h would equal the line integral between the first two line integrals.

Also, if that is right, how would you go about finding the volume between line integrals if the base line integral as at equal angles to the other two and not perpendicular to an axis? Would you have to break it into halves and do a coordinate transformation?

Edited by metacogitans

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For finding the volume between a multivariable line integral (x,y) and a single variable integral along the x axis, does this look right?

In particular I'm looking at f(x,y) parabaloids in the format z=axn+bym if that makes a difference; you can assume that f(x,y)-f(y)=f(x)

 

I'm having trouble visualising what you are trying to say or do.

A single variable function such as y = f(x) has only area, not volume.

 

Are you trying to calculate the volume using the area between y = f(x) and the x axis in the xy plane as a base and erecting perpendiculars to describe bowl shaped paraboloids in the z direction?

Edited by studiot

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okay, I think I might have put line integrals in the original post unnecessarily; I was trying to describe the boundaries of where I want to find the volume as where those line integrals 'are' on a graph.

 

I'm thinking now I should have wrote it ( Volume[0,0][x+h,0] [x+h,y+m(x+h)]f(x,y) - Volume[0,0][x,0][x,y]f(x,y) ) / h = x,0x,y f(x,y)

with 'volume' implying a volume integral beneath f(x,y) within the two points in subscript right next to the word; and m being the slope of the hypotenuse in the xy plane where z=0

 

Still don't know if that's right

 

Say there's a parabaloid z=2x2+3y2 and I want to know the volume underneath it to the xy plane where z=0, within the (x,y) boundaries (0,0) (3,0) (3,2)

Edited by metacogitans

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I think I've figured out the notation for what I was looking for now, tell me what you think:

aZ4OLyH.png

Edit: Woops, those parameters aren't listed properly, those would be the colored geometrical segments. The vertices of the section are (0,0,0)(x,0,0)(x,y,0)(x,0,z(x))(x,y,(z(y)+z(x)) and coordinates of the section follow the parameters: f(z(x)+z(y)) ≥ z ≥ 0; y ≥ 0

Edited by metacogitans

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Here we go, I think multivariable calculus finally just clicked for me:

9cgA0mE.png

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OK you are nearly there.

 

Imagine that you have a small rectangular patch, of width dx and length dy, standing on the xy plane and have erected a rectangular prism to just touch the underside of some function z = f(x,y) as shown in the sketch.

 

The volume of this prism is, of course, base area time height.

The base area is dx times dy or dxdy

The height is the average value of this function f(x, y) over the patch.

 

V = z*dx*dy=f(x,y)dxdy.

 

If we now sum these over the whole area on the xy plane we obtain the volume you seek.

 

post-74263-0-44508200-1461793197_thumb.jpg

[math]V = \int {f(x,y)dxdy} [/math]

 

Now this is a double integral and we also need to specify the limits, first as a function of x for the first integral and then as a function of y for the second integral or the other way round, depending upon which we do first.

Can you do double integrals?

The first integral will be something like from x = -5 to x = +5 ie a normal integral.

The limits of the second integral will be a function, not simple numbers, so you need a second expression say y = g(x) relating y to x in the x y plane ie for z=0.

 

.

 

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OK you are nearly there.

 

Imagine that you have a small rectangular patch, of width dx and length dy, standing on the xy plane and have erected a rectangular prism to just touch the underside of some function z = f(x,y) as shown in the sketch.

See, this is exactly the kind of question I would never touch :) And I won't even try to pin you down by asking you exactly what width dx means!!

Edited by wtf

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OK you are nearly there.

 

Imagine that you have a small rectangular patch, of width dx and length dy, standing on the xy plane and have erected a rectangular prism to just touch the underside of some function z = f(x,y) as shown in the sketch.

 

The volume of this prism is, of course, base area time height.

The base area is dx times dy or dxdy

The height is the average value of this function f(x, y) over the patch.

 

V = z*dx*dy=f(x,y)dxdy.

 

If we now sum these over the whole area on the xy plane we obtain the volume you seek.

 

attachicon.gifdblint1.jpg

[math]V = \int {f(x,y)dxdy} [/math]

 

Now this is a double integral and we also need to specify the limits, first as a function of x for the first integral and then as a function of y for the second integral or the other way round, depending upon which we do first.

Can you do double integrals?

The first integral will be something like from x = -5 to x = +5 ie a normal integral.

The limits of the second integral will be a function, not simple numbers, so you need a second expression say y = g(x) relating y to x in the x y plane ie for z=0.

 

.

 

Thanks for the reply,

I think I remembered to do most of those things but it sounds like I'm forgetting some important notation and also making up some of it, and that's not good if I want other people to be able to read it.

I checked my math by toying around with a 3d graphing calculator, and whatever I'm doing geometrically is working correctly, but algebraically the equations I typed are wrong (especially in the first two posts) or not native to calculus.

 

What do you think of this second version of the image I made though? Does it look correct more or less? Anything stick out like it needs correcting?

9cgA0mE.png

Just some notes, the green function and the green segment beneath it on the x,y axis are supposed to represent any line integral for the paraboloid you want that can be defined in terms of x and y (I kept the variables the same name as the axis for simplicity, though I color coded them) and starts at a segment on the x,y plane going through (0,0) .

 

With x and y defined, the line integral from (x,0) to (x,y) (the red segment in the image) to the paraboloid z(x,y) can be found as the integral given in the image: since value y and z(x) at value x form a rectangle, that is one piece of the line integral's area, while the second part would just be the function z(y) integrated, since leaving the x axis, it just begins to affect the dimensions of the paraboloid, and we just accounted for how z(x) affects the shape of the paraboloid along that line integral so we don't have to worry about it.

 

Then, integrating that line integral, we get an anti-derivative for finding volume.

 

I think all of that checks out unless I totally missed something.

Edited by metacogitans

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