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Find volume of a cone using integration

  • Thread starter brandy
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  • #1
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Homework Statement


Approximate this hill to a smooth cone with an elliptical base. find its volume by integration


Homework Equations


n/a?


The Attempt at a Solution


This hill is from a contour map and i have approximated the formula for the ellipse and the height.
i found the area of the ellipse and multiplied it by the height/3 but the questions says use an integration method.
help!!!
 
Last edited:

Answers and Replies

  • #2
225
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take the cone and ellipse part separately,consider a small strip in either of them,using
the equation of ellipse and cone ,(use parametric forms if necessary),integrate and add up.
 
  • #3
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the point X ( proportional to a) Z (proportional to b) and Y (proportional to h)
x/(h-y)=a/h
and
z/(h-y)=a/h

so now what?
do i rearange them both and integrate?
HELP ME!!!
im a real dummy. so real its not funny. i dont really know what monty37 is saying.
 
  • #4
Cyosis
Homework Helper
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Lets orient the cone so that the z-axis goes through the center of the ellipse and the top of the hill. This way the base of the hill lies in the xy-plane. Now slice the cone parallel to the xy-plane into a lot of ellipsis. You now want to find an expression for the surface of each ellipse. The area of an ellipse is given by [itex]\pi a b[/itex], with a the semi major axis and b the semi minor axis. Express a and b as a function of z (the height). If done correctly you will have a formula depending on z that gives the area of an ellipse at height z. If we take the base of the hill to be z=0 and the top of the hill to be z=h, we want to add all the ellipses between 0 and h together. Since it's a continuous distribution we integrate. The general formula will look like [itex]\int_0^h A(z)dz
[/itex], with A(z) the function that gives the area of the ellipse at height z.
 
  • #5
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You could also use double integration. Suppose [tex]z = f(x, y)[/tex] defines the surface of your cone. The projection onto the [tex]XY[/tex] plane is precisely the region of integration (in your case it seems to be an ellipse), so your double integral which would yield the volume you're looking for would seem to be

[tex]\displaystyle \iint_{D} z dxdy,[/tex]​

where you could use the change of variables theorem to accommodate the ellipse, since we would have

[tex]D \equiv \{(x,y) \in \mathbb{R}^2 : \frac{x^2}{a^2} + \frac{y^2}{b^2} \leq 1\}.[/tex]​

Good luck.
 
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  • #6
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Lets orient the cone so that the z-axis goes through the center of the ellipse and the top of the hill. This way the base of the hill lies in the xy-plane. Now slice the cone parallel to the xy-plane into a lot of ellipsis. You now want to find an expression for the surface of each ellipse. The area of an ellipse is given by [itex]\pi a b[/itex], with a the semi major axis and b the semi minor axis. Express a and b as a function of z (the height). If done correctly you will have a formula depending on z that gives the area of an ellipse at height z. If we take the base of the hill to be z=0 and the top of the hill to be z=h, we want to add all the ellipses between 0 and h together. Since it's a continuous distribution we integrate. The general formula will look like [itex]\int_0^h A(z)dz
[/itex], with A(z) the function that gives the area of the ellipse at height z.
The most easy way to do the sum is as shown above...

Now i havent been taught with cylindrical coordinates. And i am not used to spherical co ordinates... Cyosis, would it be possible to do the sum through cylindrical co ordinates with more ease?
 

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