How Do Integrals Behave In Higher Dimensions?

Question

As I started looking into Linear Algebra and reading Flatlands: A Romance of Many Dimensions after learning Calculus, I wondered how integrals act in higher dimensions. So here I am on Stack Exchange. How do integrals behave in higher dimensions?

Answer 1

So far you've been taking integrals with respect to one variable, whether that be $x$ or $y$. An integral's main purpose is to provide the area under the curve.

Of course in higher dimensions, for example, 3$D$, that area is known as "volume".

Now you actually have 2 axis, the $x$ and $y$ axis, and so you will encounter what's known as a double integral, where you integrate with respect to both axes.

Right now you are fimilar with single integrals, or integrals pertaining to only one axes.

Here is a visual representation.

Here you can integrate via $x$ or $y$, and you will get the area of the blue region.

In 3D, you will have two set of axes, and your output will be on the third axes (usually $z$)

Notice how here the axes are $x$ and $y$ and your output is in the $z$ axis. Here you have volumn between two sets of points.

Anyways, that's the basic idea. Integrals.

Answer 2

I assume you are asking how we can use integrals in higher dimensions.

If you had a function $f(x,y)$, you could use a double integral to find the volume underneath the surface between given boundaries.

For a really easy example, lets say that we have that $f(x,y) = 5$.This is a flat plane in 3 dimensional space raised $5$ units above the origin. Let's say we wanted to find the volume underneath this from $x=5$ to $x=10$ and $y=-2$ to $y=3$

We can calculate this using a double integral

$$\int_{-2}^3 \int_5^{10} 5 \ dx \ dy $$ $$\int_{-2}^3 \left[5x \right]_{x=5}^{10} \ dy $$ $$\int_{-2}^3 25 \ dy $$ $$[25y]_{y=-2}^3$$ $$75 - (-50) = 125$$

The volume of the space I described is $125$ cubic units.

But notice that the area I described was also simply a cube with side length $5$. The volume of this cube is $5^3 = 125$ cubic units, so this is a validation that the integral solution is correct.

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Although not so easy to visualize, we can keep this going with a triple integral to find the "hypervolume" of some sort of space in the 4th dimension with an equation $f(x,y,z)$

Answer 3

How bout this? Your velocity as a function of time is $<2t,t^2>$. Then your location is $\int <2t,t^2>\, dx=<t^2,t^3/3>+C$.