Evaluation of Triple Integrals

Section 6.2 Evaluation of Triple Integrals

As with double integrals, triple integrals can be evaluated by computing iterated integrals if the domain is given in a suitable form. The form is

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\begin{matrix} (6.2) & D = {(x, y, z) : a \leq x \leq b, φ (x) \leq y \leq Φ (x), ψ (x, y) \leq z \leq Ψ (x, y)} . \end{matrix}

Note that the projection of

D

along the

z

-axis onto the

x y

-axis is given by

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D_{0} := {(x, y, z) : a \leq x \leq b, φ (x) \leq y \leq Φ (x)},

and that the condition

ψ (x, y) \leq z \leq Ψ (x, y)

indicates that

D

is between the graph of

ψ

and

Ψ

over

D_{0} .

By slicing the domain into thin domains similarly as done in Section 5.2 one can show the following generalised version of Fubini’s theorem.

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Theorem 6.2. Fubini’s Theorem.

Suppose that

D

has the form (6.2) with

φ, Φ

continuous on

[a, b],

and

ψ, Ψ

continuous on

D_{0}

as defined above. If

f

is continuous on

D

then

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∭_{D} f (x, y, z) d x d y d z = \int_{a}^{b} (\int_{φ (x)}^{Φ (x)} (\int_{ψ (x, y)}^{Ψ (x, y)} f (x, y, z) d z) d y) d x .

Note that we can interchange the roles of

x, y, z

arbitrarily.

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Example 6.3.

Let

D

be the domain bounded below by

z = \sqrt{x^{2} + y^{2}},

and bounded from above by

9 x^{2} + 4 y^{2} + z^{2} = 5 .

Find the limits for the triple integral of a function

f (x, y, z)

over the domain

D .

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Solution.

We first find the projection,

D_{0},

of the domain onto the

x y

-plane. The intersection of

z = \sqrt{x^{2} + y^{2}}

and

9 x^{2} + 4 y^{2} + z^{2} = 5

forms a curve in space. Its projection onto the

x y

-plane is the boundary of

D_{0} .

To find that projection we eliminate

z

from the two equations. If we substitute the first into the second equation we get

9 x^{2} + 4 y^{2} + z^{2} = 9 x^{2} + 4 y^{2} + x^{2} + y^{2} = 10 x^{2} + 5 y^{2} = 5,

which we can write as

2 x^{2} + y^{2} = 1 .

The last equation describes an ellipse, so we have that

D_{0}

is given by

$- \frac{1}{\sqrt{2}} \leq x \leq \frac{1}{\sqrt{2}}$
$- \sqrt{1 - 2 x^{2}} \leq y \leq \sqrt{1 - 2 x^{2}} .$

From the conditions on

z

we finally get

$\sqrt{x^{2} + y^{2}} \leq z \leq \sqrt{5 - 9 x^{2} - 4 y^{2}}$

Hence by Fubini’s Theorem we can write

\int_{D} f (x) d x = \int_{- 1 / \sqrt{2}}^{1 / \sqrt{2}} (\int_{- \sqrt{1 - 2 x^{2}}}^{\sqrt{1 - 2 x^{2}}} (\int_{\sqrt{x^{2} + y^{2}}}^{\sqrt{5 - 9 x^{2} - 4 y^{2}}} f (x, y, z) d z) d y) d x .

f

is given we need to compute three integrals, starting with the innermost.

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