any context? bounds ? Complete problem statement ?NODARman said:Homework Statement:: .
Relevant Equations:: .
Hi, just wondering does this mean the triple integral of "y" with respect to "x"?
$$
\int \frac{d^{3} x}{y^{3}} .
$$
BvU said:any context? bounds ? Complete problem statement ?
I found it in a textbook, it's very general "equation".haruspex said:Is y a function of position in three dimensions?
What you posted is an expression, not an equation. What is the rest of it?NODARman said:I found it in a textbook, it's very general "equation".
A triple integral is a way to integrate over a three-dimensional region. It extends the concept of a single integral over a line and a double integral over an area to integration over a volume. It is often used in physics and engineering to compute quantities like mass, volume, and center of mass in three-dimensional space.
To set up a triple integral, you need to define the limits of integration for each variable. This usually involves determining the bounds for z first, then y, and finally x. The general form of a triple integral is ∫∫∫ f(x, y, z) dz dy dx, where f(x, y, z) is the function being integrated. The order of integration can vary depending on the region of integration and the function.
The order of integration in a triple integral can be dz dy dx, dy dz dx, or any other permutation of x, y, and z. The order matters because it can simplify the computation depending on the region of integration and the function being integrated. Sometimes changing the order of integration can make the integral easier to evaluate.
To change the order of integration, you need to re-evaluate the limits of integration for each variable based on the new order. This often involves sketching the region of integration and determining the new bounds for each variable. The goal is to express the same region of integration in a different order to simplify the integral.
Sure! Consider the triple integral ∫∫∫ (x + y + z) dV over the region where 0 ≤ x ≤ 1, 0 ≤ y ≤ 1, and 0 ≤ z ≤ 1. This can be set up as ∫ from 0 to 1 (∫ from 0 to 1 (∫ from 0 to 1 (x + y + z) dz) dy) dx. Evaluating the innermost integral with respect to z, we get ∫ from 0 to 1 (∫ from 0 to 1 [(xz + yz + (z^2)/2) evaluated from 0 to 1] dy) dx. This simplifies to ∫ from 0 to 1 (∫ from 0 to 1 (x + y + 1/2) dy) dx. Next, we integrate with respect to y: ∫ from 0 to