Review
These notes will supplement (not replace) the lectures based on Section 16.3 Section 16.3
(i) Double integrals over general regions: We defined double integrals over rect-angles in the last section. How does one define the double integral of a function f over a general region D? Choose an arbitrary rectangle R which contains D. Define a new function F on R as follows.
F(x, y) = f (x, y), on D = 0, outside D.
The function F is the same is f on D and zero outside D. Hence, it doesn’t matter which rectangle we choose. If F is integrable over R, we can define, for a general region D, Z Z D f(x, y)dA = Z Z R F(x, y)dA.
If the region D is “nice”, the integral on the right-hand side will exist.
(ii) Type I and II regions: If a given region D is of a specific type, it simplifies the double integral considerably.
(a) A planar region DI is of Type I if it lies between the graphs of two continuous
functions of x. That is,
DI = {(x, y)| a ≤ x ≤ b, g1(x) ≤ y ≤ g2(x)}
where g1 and g2 are continuous on [a, b].
To find RR
DIf(x, y)dA, we choose a rectangle R = [a, b] × [c, d] which contains
DI. Then, Z Z DI f(x, y)dA = Z Z R F(x, y)dA = Z b a Z d c F(x, y)dydx
Since F (x, y) = 0 for y ≥ g2(x) as well as y ≤ g1(x), and F (x, y) = f (x, y)
when g1(x) ≤ y ≤ g2(x). Thus, Z d c F(x, y)dy = Z g2(x) g1(x) f(x, y)dydx. This gives us Z Z DI f(x, y)dA = Z b a Z g2(x) g1(x) f(x, y)dydx. 1
(b) A planar region DII is of type II if it lies between the graphs of two continuous
functions of y. That is,
DII = {(x, y)| c ≤ y ≤ d, h1(y) ≤ x ≤ h2(y)} where h1 and h2 are continuous on [c, d].
Similar to above, we get Z Z DI I f(x, y)dA = Z d c Z h2(y) h1(y) f(x, y)dxdy.
Remark: Remember that in both the instances, the limits of the inner integral are functions of one variable but the integration is done with the opposite with respect to the other variable.
0.1 Example: Evaluate the double integral Z Z
D
xcos y dA, Dbounded by y = 0, y = x2, x= 1. Solution: We can write D as
D= {(x, y)| 0 ≤ x ≤ 1, 0 ≤ y ≤ x2}.
I have not drawn the region but you must do so. Clearly, D is Type I region and thus Z Z D xcos ydA = Z 1 0 Z x2 0 xcos y dydx = Z 1 0 (x sin y]x02dx = Z 1 0 xsin x2 dx= 1 2 − cos x 21 0= 1 2(1 − cos 1) 0.2 Example: Evaluate the double integral
Z Z
D
y2exydA, D= {(x, y)| 0 ≤ y ≤ 4, 0 ≤ x ≤ y}. Solution: The region D is a Type II region, hence
Z Z D y2exydA= Z 4 0 Z y 0 y2exydx dy= Z 4 0 y2 y e xy y 0 dy = Z 4 0 (yey2− y)dy = 1 2e y2 −y 2 2 4 0 = 1 2 e 16− 17 .
0.3 Example: Find the volume of the solid under the surface z = xy and above the triangle with vertices (1, 1), (4, 1) and (1, 2).
Solution: The triangle D with the given points as vertices can be seen to be a Type I region as follows. In the triangle, we see that x lies between x = 1 and x = 4 while
the y-component lies between y = 1 and the line joining the points (1, 2) and (4, 1), which has the equation y = −13x+ 73. We thus write
D= {(x, y)| 1 ≤ x ≤ 4, 1 ≤ y ≤ −x 3 +
7 3}.
Again, I have not drawn the region but you have to do so. Note that the function f(x, y) = xy is always non-negative on D. Hence, the volume V of the solid under the surface z = xy and above the given triangle D is
V = Z Z D xy dA= Z 4 1 Z −x3+73 1 xy dydx = Z 4 1 xy2 2 − x 3+ 7 3 1 dx= 1 2 Z 4 1 x −x 3 + 7 3 2 − x ! dx = 1 18 Z 4 1 x3− 14x2+ 40x dx = 1 18 x4 4 − 14x3 3 + 20x 2 4 1 = 1 18 1 4(256 − 1) − 14 3 (64 − 1) + 20(16 − 1) = 31 8 .
0.4 Example: Find the volume of the tetrahedron bounded by the coordinate planes and the plane 3x + 2y + z = 6.
Solution: The coordinate planes are given by x = 0, y = 0 and z = 0. Let’s find the vertices of the tetrahedron first. These will be the intersections of three of the four planes given. The intersection of x = 0, y = 0 and 3x + 2y + z = 6 is (0, 0, 6). Similarly, the other three vertices are (2, 0, 0), (0, 3, 0) and (0, 0, 0). Now, the given tetrahedron is a solid that lies above the triangle D in the xy plane that has vertices (0, 0), (2, 0) and (0, 3). Please draw this triangle and the tetrahedron, even though I have not done so. The line joining (0, 3) and (2, 0) is given by y = −32x+ 3. We
can write
D= {(x, y)| 0 ≤ x ≤ 2, 0 ≤ y ≤ −3 2 + 3},
which is a Type I region. Now, the volume V of the tetrahedron is the double integral of the function 6 − 3x − 2y over D.
V = Z (6 − 3x − 2y)dA = Z 2 0 Z −32x+3 0 (6 − 3x − 2y)dy dx = Z 2 0 6y − 3xy − y 2− 3 2x+3 0 dx = Z 2 0 6 −32x+ 3 − 3x −32x+ 3 − −32x+ 3 2! dx = Z 2 0 −9x + 18 + 92x2− 9x −9 4x 2 + 9x − 9 dx = Z 2 0 9 4x 2 − 9x + 9 dx= 3 4x 3 −92x2+ 9x 2 0 = 6.
0.5 Example: Find the volume of the solid enclosed by the cylinders z = x2, y = x2 and the planes z = 0, y = 4.
Solution: In these type of problems, it is not necessary to draw the given solid. You need to find the region in the xy plane above which the solid is located. In this case, we know that the solid lies above the region D bounded by the line y = 4 and the parabolay = x2 in the xy plane. In this region, x lies between −2 and 2. Hence
D= {(x, y)| − 2 ≤ x ≤ 2, x2 ≤ y ≤ 4},
which is a Type I region. Hence, the volume of the solid is the double integral over Dof the function f (x, y) = x2. V = Z Z x2dA= Z 2 −2 Z 4 x2 x2dydx = Z 2 −2 x2y4 x2dx= Z 2 −2 (4x2− x4)dx = 4 3x 3 −15x5 2 −2 = 128 15 .
(iii) Reversing the order of integration: For double integrals on general regions, we cannot simply switch the order of integration because the Fubini Theorem doesn’t directly apply on such regions. Nonetheless, it is possible that a given double integral is difficult to solve and we’d like to reverse the order of integration. For this, we first find the planar region D on which the double integral is being evaluated. Next, if Dis originally a Type I region, we can switch the order of integration by expressing it as a Type II region and evaluating the integral.
0.6 Example: Evaluate the integral by reversing the order of integration. Z 1
0
Z 3
3y
ex2dx dy.
Solution: We can see that the inner integral cannot be solved by elementary meth-ods. In order to reverse the order of integration we first find the region on which the integral is being evaluated. Looking at the limits on x and y, we can express this region D as
D= {(x, y)| 0 ≤ y ≤ 1, 3 ≤ x ≤ 3y}.
This is a Type II region. It is given by the triangle in the xy plane which is bounded by the lines y = x3, y = 0 and x = 3. I have not drawn the region but you have to do so. Hence, Z 1 0 Z 3 3y ex2dx dy= Z Z D ex2dA.
Now, to switch the order of integration, we express D as a Type I region. To do this we must express point (x, y) on D such that x lies between two real numbers
and y is between two continuous functions of x. It is clear that 0 ≤ x ≤ 3 and also 0 ≤ y ≤ x3. That is, D= {(x, y)|0 ≤ x ≤ 3, 0 ≤ y ≤ x 3}. Hence, Z 1 0 Z 3 3y ex2dx dy= Z Z D ex2dA= Z 3 0 Z x/3 0 ex2dydx = Z 3 0 yex2ix/3 0 dx= Z 3 0 1 3xe x2 dx = 1 3 Z 9 0 1 2e udu= 1 6 e 9− 1).
where we have used the substitution u = x2 in the last step.
0.7 Example: Evaluate the integral by reversing the order of integration. Z 4 0 Z 2 √x 1 y3+ 1dydx.
Solution: The double integral is taken over the region D= {(x, y)|0 ≤ x ≤ 4, √x≤ y ≤ 2},
which is a Type I region and is represented by the region above the curve y =√x, below the line y = 2 and between x = 0 and x = 4. Now, we can express this region as a Type II region as follows
D= {(x, y)| 0 ≤ y ≤ 2, 0 ≤ x ≤ y2}. Therefore, Z 4 0 Z 2 √x 1 y3+ 1dydx= Z Z D 1 y3+ 1dA= Z 2 0 Z y2 0 1 y3+ 1dxdy = Z 2 0 x y3+ 1 y2 0 dy= Z 2 0 y2 y3+ 1dy = Z 9 1 1 3 1 udu= 1 3ln 9.
where we have used the substitution u = y3+ 1 in the last step.
(iv) If a region D is the union of two regions D1 and D2. That is, D = D1∪ D2 then the
double integral over D is simply the sum of the double integrals over D1 and D2.
Z Z D f(x, y)dA = Z Z D1 f(x, y)dA + Z Z D2 f(x, y)dA.
Sometimes you won’t be able to express a given region as Type I or Type II. I such a case, see if you can write it as the union of two or more such regions.
