Finding Triple Integral: Planes at x1=0, x2=0, x3=2

[tigertan]

Okay so I need to try to find the triple integral for the following.

Planes at x₁=0, x₂=0, x₃=2
Area: x₃=x₁²+₂², x₁≥0, x₂≥0

∫_W√(x₃-x₂²dx


So I understand that need to find 3 integrals, the first being
∫₀² ∫₀^√x₃-2 ∫₀^{√x₃2-x₂2} √x₃-x₂² dx₁dx₂dx₃

I really don't understand how to find the upper and lower limits??!?

 

[jvicens]

Hello,

Thank you for your forum post. I am a scientist and I would be happy to help you find the triple integral for the given problem.

First, let's define the variables and limits of integration. We have three variables, x1, x2, and x3, and we are given the following information:

1. Planes at x1=0, x2=0, x3=2
2. Area: x3=x1^2+2^2, x1≥0, x2≥0

From the first point, we can see that the planes at x1=0 and x2=0 will form the boundaries of our integration. This means that the limits of integration for x1 and x2 will be from 0 to some value, which we will determine later.

The second point gives us the equation for the area that we need to integrate over. We can rewrite this equation as x3 = x1^2 + 4, which means that x3 is a function of x1. This will help us determine the limits of integration for x3.

Now, let's look at the integrand, which is √(x3-x2^2). We can see that x2 is also a part of the integrand, so we will need to integrate over x2 as well. This means that our final triple integral will look like this:

∫0^a ∫0^b ∫0^√(x3-x2^2) √(x3-x2^2) dx1dx2dx3

To determine the limits of integration for x3, we need to look at the equation x3 = x1^2 + 4. We know that x3=2 is a plane, so we can set x1^2 + 4 = 2 and solve for x1. This gives us two solutions, x1 = √2 and x1 = -√2. However, since x1≥0, we can only use the positive solution, which is x1 = √2. This will be our upper limit for x1.

Now, to determine the limits of integration for x2, we need to look at the equation x3 = x1^2 + 4. We can rewrite this as x2^2 = x3 - x1^2. This means that x2