Evaluation of Double Integrals

Section 5.2 Evaluation of Double Integrals

In this section we will learn how to compute a double integral over some domains. The idea is to reduce the computation to the evaluation of two integrals over functions of one variable only. We consider special domains of the form

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\begin{matrix} (5.3) & D_{1} = {(x, y) \in R^{2} : a \leq x \leq b, φ (x) \leq y \leq Φ (x)}, \end{matrix}

or the same situation with

x

and

y

switched

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\begin{matrix} (5.4) & D_{2} = {(x, y) \in R^{2} : a \leq y \leq b, φ (y) \leq x \leq Φ (y)}, \end{matrix}

where

φ

and

Φ

are continuous functions on the interval

[a, b],

and

φ (x) \leq Φ (x)

for all

x \in [a, b] .

The domain

D

is the region between the graphs of

φ

and

Φ .

Both possibilities are shown in Figure 5.10

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Figure 5.10. The two types of special domains.
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Assume now that we have a domain of the type

D_{1} .

Recall that, the double integral of a function

f

can be interpreted as the volume of the region under its graph. To get an approximation of that volume we can `cut’ the region parallel to the

y

-axis into thin slices, and then add the volumes of the slices up. More precisely, we choose a partition

a = x_{0} < x_{1} < \dots < x_{n} = b .

We then look at the slice between

x_{i - 1}

and

x_{i}

as shown in Figure 5.11.

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Figure 5.11. Cutting a domain into thin slices
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Its thickness is

Δ x_{i} := x_{i} - x_{i - 1} .

We next choose

x_{i}^{*} \in [x_{i - 1}, x_{i}] .

The approximate surface area of one side of the slice is then

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F (x_{i}^{*}) := \int_{φ (x_{i}^{*})}^{Φ (x_{i}^{*})} f (x_{i}^{*}, y) d y .

Hence the approximate volume of one of the slices is

F (x_{i}^{*}) Δ x_{i} .

Adding up we conclude that

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\sum_{i = 1}^{n} F (x_{i}^{*}) Δ x_{i}

is an approximation of the volume under the graph of

f .

The last expression is a Riemann sum for

F .

Passing to the limit as

max Δ x_{i} \to 0

the above suggests that

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\iint_{D_{1}} f (x, y) d x d y = \int_{a}^{b} (\int_{φ (x)}^{Φ (x)} f (x, y) d y) d x .

The integral

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\int_{φ (x)}^{Φ (x)} f (x, y) d y

is usually called the inner integral, the integral over

[a, b]

the outer integral, and the full expression an iterated integral. The above is not a rigorous proof, but strong evidence that a formula like the above holds. One can show that the formula is true, at least if

f, φ, Φ

are all continuous. Similar arguments can be applied with

x

and

y

interchanged, leading to a formula for domains of the form

D_{2} .

We state the result as a theorem.

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Theorem 5.12. Fubini’s Theorem.

Suppose that

φ, Φ

are continuous functions on the interval

[a, b],

and that

f

is continuous on a domain of the form

D_{1}

D_{2}

as defined in (5.3) and (5.4), respectively. Then

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\iint_{D_{1}} f (x, y) d x d y = \int_{a}^{b} (\int_{φ (x)}^{Φ (x)} f (x, y) d y) d x

and

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\iint_{D_{2}} f (x, y) d x d y = \int_{a}^{b} (\int_{φ (y)}^{Φ (y)} f (x, y) d x) d y .

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Example 5.13.

Let

D

be the domain below

y = 2 x

and above

y = x^{2} .

Write

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\int_{D} f (x) d x

as an iterated integral.

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Solution.

We first determine where

2 x \geq x^{2} .

To do so we solve the equation

2 x = x^{2} .

One solution certainly is

x = 0 .

x \neq 0

we can divide by

x

and get

2 = x,

providing another solution. Hence we have

2 x \geq x^{2}

for

0 \leq x \leq 2 .

Make a sketch of the domain to visualise the situation. We therefore have

a = 0,

b = 2,

φ (x) = x^{2}

and

Φ (x) = 2 x .

Hence, by Fubini’s Theorem

\int_{D} f (x) d x = \int_{0}^{2} (\int_{x^{2}}^{2 x} f (x, y) d y) d x,

which is an iterated integral as required.

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Example 5.14.

Compute

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\iint_{D} x^{2} y d x d y,

where

D

is the domain from Example 5.13.

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Solution.

By the result from the previous example we have

\iint_{D} f (x, y) d x d y = \int_{0}^{2} (\int_{x^{2}}^{2 x} x^{2} y d y) d x .

We first evaluate the inner integral:

\int_{x^{2}}^{2 x} x^{2} y d y = x^{2} \int_{x^{2}}^{2 x} y d y = x^{2} \frac{y^{2}}{2} |_{x^{2}}^{2 x} = \frac{x^{2}}{2} ((2 x)^{2} - (x^{2})^{2}) = 2 x^{4} - \frac{x^{6}}{2} .

Then we evaluate the outer integral:

\begin{aligned} \iint_{D} f (x, y) d x d y & = \int_{0}^{2} (\int_{x^{2}}^{2 x} x^{2} y d y) d x \\ = \int_{0}^{2} 2 x^{4} - \frac{x^{6}}{2} d x \\ = {\frac{2 x^{5}}{5} |}_{0}^{2} - {\frac{x^{7}}{2 \cdot 7} |}_{0}^{2} \\ = \frac{2^{6}}{5} - 0 - \frac{2^{7}}{2 \cdot 7} + 0 \\ = \frac{7 \cdot 2^{6} - 5 \cdot 2^{6}}{35} = \frac{2 \cdot 2^{6}}{35} = \frac{2^{7}}{35}, \end{aligned}

which is the required result.

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