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**15.2** Double Integrals over

### General Regions

### Double Integrals over General Regions

For single integrals, the region over which we integrate is always an interval.

But for double integrals, we want to be able to integrate a
*function f not just over rectangles but also over regions D of *
more general shape, such as the one illustrated in Figure 1.

**Figure 1**

### Double Integrals over General Regions

*We suppose that D is a bounded region, which means that *
*D can be enclosed in a rectangular region R as in Figure 2. *

*Then we define a new function F with domain R by*

**Figure 2**

### Double Integrals over General Regions

**If F is integrable over R, then we define the double ****integral of f over D by**

*Definition 2 makes sense because R is a rectangle and so *

### ∫∫

_{R}*F(x, y) dA has been previously defined.*

### Double Integrals over General Regions

The procedure that we have used is reasonable because
*the values of F(x, y) are 0 when (x, y) lies outside D and so *
they contribute nothing to the integral.

*This means that it doesn’t matter what rectangle R we use *
*as long as it contains D.*

*In the case where f(x, y) ≥ 0, we can still interpret *

### ∫∫

_{D}*f(x, y) dA as the volume of the solid that lies above D*

*and under the surface z = f(x, y) (the graph of f).*

### Double Integrals over General Regions

You can see that this is reasonable by comparing the

*graphs of f and F in Figures 3 and 4 and remembering that *

### ∫∫

_{R}*F(x, y) dA is the volume under the graph of F.*

**Figure 3** **Figure 4**

### Double Integrals over General Regions

*Figure 4 also shows that F is likely to have discontinuities *
*at the boundary points of D.*

*Nonetheless, if f is continuous on D and the boundary *
*curve of D is “well behaved”, then it can be shown that *

### ∫∫

_{R}*F(x, y) dA exists and therefore*

### ∫∫

_{D}*f(x, y) dA exists.*

**In particular, this is the case for type I and type II regions.**

### Double Integrals over General Regions

**A plane region D is said to be of type I if it lies between the ***graphs of two continuous functions of x, that is,*

*D = {(x, y) | a ≤ x ≤ b, g*_{1}*(x) ≤ y ≤ g*_{2}*(x)}*

*where g*_{1} *and g*_{2} *are continuous on [a, b]. Some examples *
of type I regions are shown in Figure 5.

Some type I regions

**Figure 5**

### Double Integrals over General Regions

In order to evaluate

### ∫∫

_{D}*f(x, y) dA when D is a region of*

*type I, we choose a rectangle R = [a, b] × [c, d] that*

*contains D, as in Figure 6, and we let F be the function *

*given by Equation 1; that is, F agrees with f on D and F is 0 *
*outside D.*

**Figure 6**

### Double Integrals over General Regions

Then, by Fubini’s Theorem,

*Observe that F(x, y) = 0 if y < g*_{1}*(x) or y > g*_{2}*(x) because *
*(x, y) then lies outside D. Therefore*

*because F(x, y) = f(x, y) when g*_{1}*(x) ≤ y ≤ g*_{2}*(x).*

### Double Integrals over General Regions

Thus we have the following formula that enables us to evaluate the double integral as an iterated integral.

The integral on the right side of (3) is an iterated integral,
*except that in the inner integral we regard x as being *

*constant not only in f(x, y) but also in the limits of *
*integration, g*_{1}*(x) and g*_{2}*(x).*

### Double Integrals over General Regions

**We also consider plane regions of type II, which can be **
expressed as

*D = {(x, y) | c ≤ y ≤ d, h*_{1}*(y) ≤ x ≤ h*_{2}*(y)}*

*where h*_{1} *and h*_{2} are continuous. Two such regions are
illustrated in Figure 7.

Some type II regions

### Double Integrals over General Regions

Using the same methods that were used in establishing (3), we can show that

### Example 1

Evaluate

### ∫∫

_{D}*(x + 2y) dA, where D is the region bounded by*

*the parabolas y = 2x*

^{2}

*and y = 1 + x*

^{2}.

Solution:

*The parabolas intersect when 2x*^{2} *= 1 + x*^{2}*, that is, x*^{2} = 1,
*so x = ±1.*

*We note that the region D, *
sketched in Figure 8, is a

type I region but not a type II region and we can write

*D = {(x, y) | –1* *≤ x ≤ 1, 2x*^{2} *≤ y ≤ 1 + x*^{2}}

**Figure 8**

*Example 1 – Solution*

*Since the lower boundary is y = 2x*^{2} and the upper
*boundary is y = 1 + x*^{2}, Equation 3 gives

cont’d

*Example 1 – Solution*

_{cont’d}

### Properties of Double Integrals

### Properties of Double Integrals

We assume that all of the following integrals exist. For
*rectangular regions D the first three properties can be *

proved in the same manner. And then for general regions the properties follow from Definition 2.

*where c is a constant*
*If f(x, y) ≥ g(x, y) for all (x, y) in D, then*

### Properties of Double Integrals

The next property of double integrals is similar to the property of single integrals given by the equation

*If D = D*_{1} *U D*_{2}*, where D*_{1} *and D*_{2}
don’t overlap except perhaps on
their boundaries (see Figure 17),

then ^{Figure 17}

### Properties of Double Integrals

Property 9 can be used to evaluate double integrals over
*regions D that are neither type I nor type II but can be *

expressed as a union of regions of type I or type II.

Figure 18 illustrates this procedure.

**Figure 18**

*(a) D is neither type I nor type II.* *(b) D = D*_{1} *∪ D*_{2}*, D*_{1}*is type I, D*_{2} is type II.

### Properties of Double Integrals

The next property of integrals says that if we integrate the
*constant function f(x, y) = 1 over a region D, we get the *
*area of D:*

### Properties of Double Integrals

Figure 19 illustrates why Equation 10 is true: A solid
*cylinder whose base is D and whose height is 1 has *

*volume A(D) * *1 = A(D), but we know that we can also write *
its volume as

### ∫∫

_{D}

^{1 dA.}*Cylinder with base D and height 1*

**Figure 19**

### Properties of Double Integrals

Finally, we can combine Properties 7, 8, and 10 to prove the following property.

### Example 6

Use Property 11 to estimate the integral

### ∫∫

_{D}*e*

*sin x cos y*

*dA,*

*where D is the disk with center the origin and radius 2.*

Solution:

Since –1 ≤ sin x ≤ 1 and –1 ≤ cos y ≤ 1, we have –1 ≤ sin x cos y ≤ 1 and therefore

*e*^{–1} *≤ e**sin x cos y* *≤ e*^{1} *= e*

*Thus, using m = e*^{–1} *= 1/e, M = e, and A(D) = *π(2)^{2} in
Property 11, we obtain