Volumes of Irregular Shapes by Integration

[labeattie]

Hi all,

Is it possible to determine the volume of a shape where, in the x and y dimensions, the shape is described by an equation, and then its elevation is described by another equation?

An example would be a parabola in the x-y plane whose elevation is based on another parabolic function. For example let's say y=x^2+4 from x=-2 to x=2 with an elevation determined by a circle of radius 4 (so zero elevation at the apex and the x-axis). Meaning z=\sqrt{16-y^2} (independent of x).

Can I get this volume by integration, or would I just need to apply some kind of finite element approach? Thanks in advance; most all of the volume integrals I could find online or in a text are just volumes of revolution.

 

[DivergentSpectrum]

yes, in order to do this it requires that the bounds of integration be functions instead of numbers.

\int_{x_{0}}^{x_{1}}\int_{y_{0}(x)}^{y_{1}(x)}z(x,y) dydx

here y₁(x) is greater than y₀(x) throughout the interval where x₀<x<x₁
look up "double integrals" for more.

also, given your problem it would be far easier to use a change of coordinates(most likely polar coordinates)

 

[labeattie]

Ah yes. I knew I was missing something (it requires that the bounds of integration be functions instead of numbers). I was having trouble making that mental leap. And I will look into the polar coordinates. I hadn't thought of those for a few years :). Thanks!

 

[HallsofIvy]

Hi all,

Is it possible to determine the volume of a shape where, in the x and y dimensions, the shape is described by an equation, and then its elevation is described by another equation?

An example would be a parabola in the x-y plane whose elevation is based on another parabolic function. For example let's say y=x^2+4 from x=-2 to x=2 with an elevation determined by a circle of radius 4 (so zero elevation at the apex and the x-axis). Meaning z=\sqrt{16-y^2} (independent of x).

The problem here is that this is not a solid at all. With just the information that "y= x^2+ 4", (x, y) is restricted to that parabola and adding z (height) just gives a "wall" which has area, not volume. Perhaps you meant to add another boundary and have x and y inside the bounded region? Say "y between y= x^2+ 4 and y= 12- x^2. Those two curves intersect at (-2, 8) and (2, 8). With height \sqrt{16- y^2}, the volume is given by \int_{x= -2}^2\int_{y= x^2+ 4}^{12- x^2}\sqrt{16- y^2} dy dx

Can I get this volume by integration, or would I just need to apply some kind of finite element approach? Thanks in advance; most all of the volume integrals I could find online or in a text are just volumes of revolution.