# Volumes of Irregular Shapes by Integration

## Main Question or Discussion Point

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 lets 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.

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)

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
Homework Helper
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 lets 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.