Can Single, Double, and Triple Integrals Vary in Variable Count?

[Calpalned]

Homework Statement

Can a single integral be used to solve a multi-variable equation, and can a triple integral be used to find the area under an y = f(x) curve?

What I'm getting at is whether or not single, double and triple integrals must be integrated with respect to their corresponding number of variables. My textbook shows single $\int f(x) dx$, double $\int \int f(x,y) dy dx$ and triple $\int \int \int f(x, y, z) dy dx dz$ but it never shows $\int f(x,y)$ nor $\int \int \int f(x, y) dy dx dz$

Homework Equations

n/a

The Attempt at a Solution

This isn't exactly a homework question, but I just want to understand the concept of integrals better.

 

[Mark44]

Homework Statement

Can a single integral be used to solve a multi-variable equation

"Solve" and "equation" aren't the right words here - you don't "solve" an integral, you evaluate it. And in place of "equation", "function" would be an appropriate choice. When you evaluate an iterated integral, in the inner integrations you are integrating a multi-variable function along one axis to, eventually, get down to an expression in one variable that you can integrate.

, and can a triple integral be used to find the area under an y = f(x) curve?

I suspose you could, but it seems like a wasted effort. You can use a double integral to find the area under a curve, y = f(x). For example, these two integrals produce the same value:
$$\int_0^1 x^2 dx$$
and
$$\int_0^1 \int_{y = 0}^{x^2}~1~dy~dx$$

You could turn this into a triple integral to get a volume that is numerically equal to the area of the preceding integrals.

What I'm getting at is whether or not single, double and triple integrals must be integrated with respect to their corresponding number of variables. My textbook shows single $\int f(x) dx$, double $\int \int f(x,y) dy dx$ and triple $\int \int \int f(x, y, z) dy dx dz$ but it never shows $\int f(x,y)$ nor $\int \int \int f(x, y) dy dx dz$

Homework Equations

n/a

The Attempt at a Solution

This isn't exactly a homework question, but I just want to understand the concept of integrals better.

 

[Calpalned]

" in place of "equation", "function" would be an appropriate choice

Is this the difference between a function and an equation:
equation $10 = 5 + x$ and $22 = x^2 + xy + z$
function $f(x) = 5+ x$ and $f(x, y, z) = x^2 + xy + z$

 

[Mark44]

Is this the difference between a function and an equation:
equation $10 = 5 + x$ and $22 = x^2 + xy + z$
function $f(x) = 5+ x$ and $f(x, y, z) = x^2 + xy + z$

More or less, as far a functions are concerned. In your first function example f maps a number x in its domain to a number 5 + x in its range. f(x) represents the "output" value for an input x value.

In your second function example, f maps an ordered triple (x, y, z) in R³ to the real number x² + xy + z.

 

[jbstemp]

I think technically you could evaluate those types of integrals, but you won't really get any thing with meaningful results.

 

[fourier jr]

The single integral $\int^{\infty}_{0}e^{-x^{2}}dx$ can't be evaluated by finding an antiderivative but can still be worked out by introducing a second integral $\int^{\infty}_{0}\int^{x}_{0}e^{-x^{2}}\,dy\,dx$ I guess it still isn't quite the same thing as working out the single integral because you get a different answer. I'm just trying to think of an integral analogy with something I saw in concrete mathematics for doing finite sums.

 

[LCKurtz]

The single integral $\int^{\infty}_{0}e^{-x^{2}}dx$ can't be evaluated by finding an antiderivative but can still be worked out by introducing a second integral $\int^{\infty}_{0}\int^{x}_{0}e^{-x^{2}}\,dy\,dx$ I guess it still isn't quite the same thing as working out the single integral because you get a different answer. I'm just trying to think of an integral analogy with something I saw in concrete mathematics for doing finite sums.

I think you mean $\int_0^\infty\int_0^\infty e^{-x^2-y^2}~dydx$, which gives the square of that integral, and which is usually evaluated by changing to polar coordinates.

 

[fourier jr]

I think that's what I meant also

 

[HallsofIvy]

Do you understand that \int_a^b \int_c^d f(x,y,z) dxdy would be a function of z?

 

[fourier jr]

yeah if you were to integrate a 3rd time it would be wrt z I suppose.