# Upper and lower bounds of a triple integral

• TheSodesa
In summary, the problem asks for finding different ways to write the integral \iiint_T f(x,y,z)dV, where T is a region bounded by the coordinate planes and surfaces y = sqrt(x) and z = 1-y in the first octant. The three different cases for the iterated integrals are dydxdz, dxdydz, and dzdxdy, with the respective upper and lower bounds of integration being (0, (1-z)^2), (0, 1-z), and (0, y^2) for x, (0, 1) for y, and (0, 1-y) for z.
TheSodesa

## Homework Statement

Let ##T \subset R^3## be a set delimited by the coordinate planes and the surfaces ##y = \sqrt{x}## and ##z = 1-y## in the first octant.

Write the intgeral
$$\iiint_T f(x,y,z)dV$$
as iterated integrals in at least 3 different ways.

## Homework Equations

$$\iiint_T f(x,y,z)dV = \int_a^b \int_{y_1 (x)}^{y_2 (x)} \int_{z_1 (x,y)}^{z_2 (x,y)} f(x,y,z) dx dy dz$$

## The Attempt at a Solution

My interpretation of this problem is that they want me to "simply" find out the upper and lower bounds of integration for 3 different cases, since there is no specific function to integrate.

This is a bit difficult to explain, since I can't show you a picture of the volume because I can't seem to be able to draw the ##y = \sqrt{x}## correctly in Matlab using this code:

syms x y z

f(x,z) = sqrt(x);
g(x,y) = 1 - y;

hold on
xlabel('x')
ylabel('y')
zlabel('z')
axis equal

ezsurf(f,0:1);
ezsurf(g,0:1)

Anyways, I'm having trouble seeing what the upper bounds are for the innermost iteration, whichever way I try and integrate this thing. For example, if I were to integrate first with respect to y, then x and finally z, I would write:
$$\int_0^1 \int_{0}^{(-z+1)^2} \int_{0}^{?} f(x,y,z) dy dx dz$$
Just looking at the horrible picture that I drew on paper (which kind of looks like a tetrahedron cut off by the ##\sqrt{x}##-surface) is not helping. Does anybody have any hints on where to start. My brain is telling me that ##?## should be a plane of some sort, but the fact that it's cut off by the ##\sqrt{x}## is really throwing me off.

EDIT: The shape is simple (as in x-, y- and z-simple) in all 3 dimensions, so I was able to print out the projections to the different coordinate planes:

y(x):

z(y)

x(z):

Last edited:
$$\int_0^1 \int_{0}^{(1-z)^2} \int_{\sqrt{x}}^{1} f(x,y,z) dy dx dz$$
I'm saying this, because
\begin{cases}

y = \sqrt{x} \iff x = y^2\\
y+z = 1 \iff z = 1- \sqrt{x} \iff x = (1-z)^2
\end{cases}

How does this look?

Last edited:
Using the same logic as above for two other cases:

##dxdydz##:

$$\int_{0}^{1} \int_{0}^{1-z} \int_{0}^{y^2}f(x,y,z) dxdydz$$
and ##dzdxdy##:
$$\int_{0}^{1} \int_{0}^{y^2} \int_{0}^{1-y} f(x,y,z) dzdxdy$$

Last edited:

## 1. What are upper and lower bounds in a triple integral?

Upper and lower bounds in a triple integral refer to the limits of integration for each variable in the integral. The upper bound is the maximum value that a variable can take, while the lower bound is the minimum value. These bounds define the region over which the integral is evaluated.

## 2. How do you determine the upper and lower bounds for a triple integral?

The upper and lower bounds for a triple integral are typically determined by examining the limits of the three variables involved in the integral. This could be based on the geometry of the region being integrated over or other constraints given in the problem.

## 3. Can the upper and lower bounds of a triple integral be negative?

Yes, the upper and lower bounds of a triple integral can be negative. This depends on the problem and the region being integrated over. The bounds can be any real numbers, including negative numbers.

## 4. What happens if the upper and lower bounds of a triple integral are switched?

If the upper and lower bounds of a triple integral are switched, the resulting integral will have a negative value. This is because the order of the bounds affects the direction of integration, which can result in a change in sign for the integral. It is important to carefully consider the order of the bounds when evaluating a triple integral.

## 5. Are there any techniques for simplifying the calculation of upper and lower bounds in a triple integral?

Yes, there are some techniques that can be used to simplify the calculation of upper and lower bounds in a triple integral. These include using symmetry to reduce the number of bounds, changing variables to make the bounds more manageable, and breaking the integral into smaller pieces to make the calculation more manageable.

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