Triple Integrals in Cartesian

Section 4.1 Triple Integrals in Cartesian

To find the volume between a surface and the

x y

-plane, a double integral partitions the domain (

x y

-plane) into squares which are the bases for cuboids (rectangular prisms). To calculate the same volume a triple integral partitions the

x,

y,

and

z

axes. The result divides a volume into cubes. See Figure 4.1.1.

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Figure 4.1.1. Triple Integral

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For double integrals the cuboids’ (rectangular prisms) volumes are height times width times depth which become the

f (x, y) \times d x \times d y .

The domain for the shape determines the limits of integration. For triple integrals the cubes’ volumes are

d x \times d y \times d z .

The domain for the shape determines the limits of integration including the vertical. As a result the process of setting up triple integrals is the same as double integrals, but there is a third pair of limits of integration. Watch the video below for an illustration of setting up a triple integral for volume.

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Example 4.1.2.

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Find the volume of the solid between the surface

z = 5 - 2 x^{2} - 3 y^{2}

and the xy-plane (

z = 0

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First we determine what function we will integrate. For a volume calculation we integrate

1 d x d y d z .

See later examples for why we would integrate something else.

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The next thing we need determine is the region over which we will be integrating. These become the bounds of the integral.

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We know the shape of both parts, but need to determine how they connect, that is, where the two surfaces intersect. Both are solved in terms of

z,

so we can simply setup up the equation

0 = 5 - 2 x^{2} - 3 y^{2} .

Note this is an ellipse in terms of

x

and

y .

It will define the x and y bounds of the integral. Because the

z = 0

is the lowest possible

z

value, that is the bottom of the shape and the other surface is the top.

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Due to symmetries of this shape in

x

and

y

and that the vertical always has the same top and bottom, we can chose to integrate in any order. We choose x then y then z.

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Because the plane is on the bottom (a

z

property) the bounds for

x

and

y

are determined solely by the surface. We can solve the surface equation for x.

\begin{aligned} z = & 5 - 2 x^{2} - 3 y^{2} \\ 2 x^{2} = & 5 - 3 y^{2} - z \\ x^{2} = & \frac{5 - 3 y^{2} - z}{2} \\ x = & \pm \sqrt{\frac{5 - 3 y^{2} - z}{2}} \end{aligned}

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Two solutions makes sense because we need both a back (the negative in this case) and a front (the positive in this case). The first part of the integral will be

\int_{- \sqrt{\frac{5 - 3 y^{2} - z}{2}}}^{\sqrt{\frac{5 - 3 y^{2} - z}{2}}} d x

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For the y bounds we need all possible y values without respect to the x value, that is at each height (z value) what the y values can be. This means we need the smallest and biggest y values. In this case we know the cross section is an ellipse and so the biggest and smallest y values occur when

x = 0 .

Using this we can solve the surface equation for the y bounds.

\begin{aligned} z = & 5 - 2 0^{2} - 3 y^{2} \\ 3 y^{2} = & 5 - z \\ y^{2} = & \frac{5 - z}{3} \\ y = & \pm \sqrt{\frac{5 - z}{3}} \end{aligned}

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Again two solutions makes sense because we need both a left and a right bound. Our integral is now

\int_{- \sqrt{\frac{5 - z}{3}}}^{\sqrt{\frac{5 - z}{3}}} \int_{- \sqrt{\frac{5 - 3 y^{2} - z}{2}}}^{\sqrt{\frac{5 - 3 y^{2} - z}{2}}} d x d y

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Finally we determine the z bounds. In this case we know the plane

z = 0

is always the bottom and the surface determines the top. We need all possible z values without respect to x or y. In this case we can see that the highest value occurs in the center (