# Triple Integrals: Finding Volume of Solid S Bounded by Planes

• MHB
• WMDhamnekar
In summary, a triple integral is a calculus concept used to find the volume of a three-dimensional solid. To set up a triple integral, the limits of integration for each variable and the function must be identified. The formula for finding the volume is V = ∭ f(x, y, z) dV. This method can also be used for solids bounded by curved surfaces, but the order of integration should be carefully chosen for easier computation.
WMDhamnekar
MHB
Find the volume V of the solid S bounded by the three coordinate planes, bounded above by the plane x + y + z = 2, and bounded below by the z = x + y.

How to answer this question using triple integrals? How to draw sketch of this problem here ?

I like your idea of sketching the graphs. Maybe start by working out the intercepts with the co-ordinate axes.

E.g. for x + y + z = 2, the x intercept is x + 0 + 0 = 2 => x = 2.

With the three intercepts it's pretty easy to draw the plane.

Prove It said:
I like your idea of sketching the graphs. Maybe start by working out the intercepts with the co-ordinate axes.

E.g. for x + y + z = 2, the x intercept is x + 0 + 0 = 2 => x = 2.

With the three intercepts it's pretty easy to draw the plane.
I computed the answer -8 as follows but it is wrong. Where am I wrong?

$\displaystyle\int_0^2\displaystyle\int_0^2\displaystyle\int_{x+y}^{2-x-y}1 dzdydx=-8$

Even if I change the upper limit of integration of y as 2-x, the answer would be $\displaystyle\int_0^2\displaystyle\int_0^{2-x} (2-2x-2y) dydx = -\frac43$ which is also wrong.

Last edited:
One thing that may be causing confusion is that the problem is not stated very well. We are told the figure is bounded by the three coordinate planes, by the plane x+ y+ z= 2, and by the plane z= x+ y. But there is NO region bounded by those five planes! The plane z= x+ y "covers" the plane z= 0. I think what is intended is the region bounded by the two coordinate planes, x= 0 and y= 0, and by x+ y+ z= 2 and z= x+ y but not by the plane z= 0. Adding x+ y+ z= 2 and x+ y- z= 0 we get 2x+ 2y= 2 or x+ y= 1. The last two planes intersect at the line x+ y= 1, z= 1. The first plane, x+ y+ z, intersects the coordinate planes in a triangle with vertices (0, 2, 0), (2, 0, 0) and (0, 0, 2). The sides have equations x= 0, y+ z= 2, y= 0, x+ z= 2, and z= 0, x+ y= 2.

The second triangle passes through the origin, (0, 0, 0), and cuts the first plane in the line x+ y= 1, z= 1.
The largest value x take in the region is 1 so we must take x from 0 to 1. For every x, y goes from 0 to 1- x, and for every x, y, z goes from z= 2- x- y.

The integral can be written a $\int_{x= 0}^1\int_{y= 0}^{1- x}\int_{z= 0}^{2- x- y}dzdxdy$.

Final answer to this question is $\displaystyle\int_{x=0}^1\displaystyle\int_{y=0}^{1-x}\displaystyle\int_{z=x+y}^{z=2-x-y} dzdydx= \frac13$

That's what I get!

## 1. What is a triple integral?

A triple integral is a mathematical concept used in multivariable calculus to find the volume of a three-dimensional object. It involves integrating a function over a three-dimensional region bounded by three planes.

## 2. How do you set up a triple integral?

To set up a triple integral, you first need to determine the limits of integration for each variable. This is done by identifying the points where the planes intersect and using those points as the boundaries for each variable. Then, you need to determine the integrand, which is the function being integrated. This function will typically involve all three variables.

## 3. What is the difference between a triple integral and a regular integral?

A triple integral involves integrating over a three-dimensional region, while a regular integral involves integrating over a one-dimensional region. In a regular integral, the limits of integration are typically constant values, whereas in a triple integral, the limits can vary for each variable.

## 4. How do you find the volume of a solid using a triple integral?

To find the volume of a solid using a triple integral, you need to set up the integral with the correct limits of integration and integrand. Then, you can solve the integral using techniques such as substitution or integration by parts. The resulting value will be the volume of the solid bounded by the given planes.

## 5. What are some real-world applications of triple integrals?

Triple integrals have many real-world applications, including calculating the mass and center of mass of three-dimensional objects, finding the volume of a fluid in a container, and determining the electric potential of a three-dimensional charge distribution. They are also used in engineering and physics to solve problems related to fluid flow, heat transfer, and electromagnetism.

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