is called an iterated integral of f over R.
Similarly, if R is a simple region defined by c ≤ y ≤ d and h1(y) ≤ x ≤ h2(y), where h1(y) and
h2(y) are continuous functions on [c, d] and let f (x, y) be a function defined on R. Then
JR = ˆ d c ˆ h2(y) h1(y) f (x, y)dx ! dy is called an iterated integral of f over R.
Notation. Sometimes IR may be written as
IR = ˆ b a ˆ g2(x) g1(x) f (x, y)dy ! dx = ˆ b a ˆ g2(x) g1(x) f (x, y) dy dx = ˆ b a dx ˆ g2(x) g1(x) f (x, y) dy, and JR may be written as
JR = ˆ d c ˆ h2(y) h1(y) f (x, y)dx ! dy = ˆ d c ˆ h2(y) h1(y) f (x, y) dx dy = ˆ d c dy ˆ h2(y) h1(y) f (x, y) dx. Example 1.1. Calculate both iterated integrals IRand JRof the function f (x, y) = 3xy2−2x2y
over R where R is bounded by x = 0, x = 2, y = 1 and y = 2.
Theorem 1.1. Suppose that f (x, y) is continuous on R, where R is defined by a ≤ x ≤ b, g1(x) ≤ y ≤ g2(x)
or
c ≤ y ≤ d, h1(y) ≤ x ≤ h2(y).
Then the double integral of f over R, denoted by ¨ R f (x, y)dA, is ¨ R f (x, y) dA = ˆ b a ˆ g2(x) g1(x) f (x, y) dy dx = ˆ d c ˆ h2(y) h1(y) f (x, y) dx dy.
Properties. Let f (x, y) and g(x, y) be two integrable functions over R, a and b be two arbitrary constants, and R = R1∪ R2. Then
(i) ¨ R af (x, y) + bg(x, y) dA = a ¨ R f (x, y) dA + b ¨ R g(x, y) dA. (i) ¨ R f (x, y) dA = ¨ R1 f (x, y) dA + ¨ R1 f (x, y) dA.
Example 1.2. (1) If R = {(x, y)| 0 ≤ x ≤ 2 and 1 ≤ y ≤ 4}, evaluate ¨
R
(6x2 + 4xy3) dA.
(2) Let R be the region bounded by y = x, y = 0 and x = 4, write ¨ R f (x, y) dA as an iterated integral. (3) Evaluate ¨ R
x dA, where R is the portion of disk x2+ y2 ≤ 16 in the second quadrant.
(4) Write ¨
R
f (x, y) dA as an iterated integral, where R is the region bounded by x = y2 and x = 2 − y.
(5) Let R be the region bounded by the graphs of y = x, y = 0 and x = 4. Evaluate ¨
R
(4ex2 − 5 sin y)dA.
(6) Let R be the region bounded by the graphs of y =¨ √x, x = 0 and y = 3. Evaluate
R
(2xy2+ 2y cos x) dA.
(7) Write ¨
R
f (x, y) dA as an iterated integral, where R is the region bounded by x2+y2 = 1 and x2+ 2y2 = 1.
(4) 0 x2 f (x, y) dy dx. (5) ˆ e 1 ˆ ln y 0 f (x, y) dx dy. (6) ˆ 3 0 ˆ x 0 f (x, y) dy dx + ˆ 6 3 ˆ 6−x 0 f (x, y) dy dx. Example 1.4. Evaluate the integral.
(1) ˆ 1 0 ˆ 1 y ex2dxdy. (2) ˆ 1 0 ˆ 1 √ x ey3dy dx. (3) ˆ 9 1 ˆ 3 √ y ex2−2x x + 1 dx dy. (4) ¨ R
x(x − 1)exydA where R is bounded by the lines x = 0, y = 0 and x + y = 2.
(5) ¨
R
(x + y) dA where R is bounded by the lines y = 2x, x = 0 and y = 4.
(6) ˆ 8 2 ˆ ln x 0 eydy dx. (7) ˆ π 0 ˆ 1+cos y 0 x2sin y dx dy. (8) ˆ 1 0 ˆ 1 0 y2 x2 + 1dy dx. (9) ˆ 1 0 ˆ 1 0 x (1 + x2+ y2)32 dx dy. (10) ˆ 2 1 ˆ x 1 x y2 x2 dy dx. 2. Area
The area of a bounded region in the real plane can be computed by ¨
R
For example, if R has the shape then, A(R) = ˆ b a ˆ g2(x) g1(x)
dy dx. If R has the shape
then, A(R) = ˆ d c ˆ h2(y) h1(y) dx dy.
Example 2.1. Find the area of the region R where R is bounded by (1) y = x and y = x2 in the first quadrant.
(2) y = x2 and y = x + 2. (3) xy = 1, y = x, x = e. (4) y = 1 4x 2− 1 and y = 2 − x. (5) x = y2, y − x = 3, y = −3, and y = 2.
Example 2.2. Let R1 = {(x, y)| x2+ 2y2 ≤ 1}, R2 = {(x, y)| 2x2+ y2 ≤ 1} and R = R1∩ R2.
Find the area of R.
3. Volume The double integral
¨
D
f (x, y)dA of a positive function f (x, y) can be interpreted as the volume under the surface z = f (x, y) over the region D.
If f (x, y) ≥ g(x, y) for all (x, y) in D, then the double integral ¨
D
(f (x, y) − g(x, y))dA is the volume between the surface z = f (x, y) and the surface z = g(x, y).
Example 3.1. For example, the volume of the tetrahedron bounded by the plane 2x+y +z = 2 and the three coordinate planes (See the figure below)
is given by V = ˆ 1 0 ˆ 2−2x 0 (2 − 2x − y) dy dx
Example 3.2. Find the volume of the solid lying in the first octant and bounded by the graphs of z = 4 − x2, x + y = 2, x = 0, y = 0 and z = 0.
Example 3.3. (1) Find the volume of the solid bounded by the graphs of z = 2, z = x2+1, y = 0 and x + y = 2.
(2) A solid T is bounded by the coordinate planes, the plane z = x + 2y + 1 and the planes x = 1 and y = 2. Find the volume of T .
(3) A solid T lies in the first octant and is bounded by the coordinate planes, and the planes x + 2y = 2 and x + 4y + 2z = 8. Find the volume of T .
(4) A solid T is bounded by the elliptic paraboloid z = 1 4x 2 + 1 9y 2
, the coordinate planes and the planes x = −2 and y = 3. Find the volume of T .
(5) A solid T lies in the first octant and is bounded by the elliptic cylinder 4x2+ z2 = 1, the plane y = x, and the planes y = 0 and z = 0.
(6) Find the volume enclosed between the two surfaces z = x2+ 3y2 and z = 8 − x2− y2.
(7) A solid T is bounded above by z = 2 − x2 − y2 and below by z = x2 + y2. Find the
volume of T .
(8) A solid T is bounded by the sphere x2+y2+z2 = 4 and the paraboloid x2+y2 = 4(1−z).
Find the volume of T .
(9) Find the volume V of the solid region T where T is bounded by the sphere x2+y2+z2 = 4
and the paraboloid x2+ y2 = 4(1 − z).
(10) A solid T is bounded by the paraboloids x2 + y2 − z = 0 and x2+ y2+ 2z = 1. Find
the volume of T .
(11) Find the volume of the solid in the first octant bounded by the circular paraboloid z = x2+ y2, the cylinder x2+ y2 = 4 and the coordinate planes. Find the volume of T . (12) A solid T is bounded by the cone z2 = xy and the plane x + y = 4. Find the volume of
T .
(13) Use double integration to find the volume of the tetrahedron bounded by the coordinate planes and the plane 3x + 6y + 4z − 12 = 0.
(14) Find the volume of the solid lying in the first octant and bounded by z = 4 − x2,
x + y = 2, x = 0 and z = 0.
(15) Find the volume enclosed between the two surfaces z = x2+ y2 and z = 8 − x2 − y2.
4. Change of Variables in Double Integrals
A transformation T from the uv-plane to the xy-plane is a function that maps points in the uv-plane to points in the xy-plane, so that
T (u, v) = (x, y),
where x = g(u, v) and y = h(u, v), for some functions g and h. We consider changes of variables in double integrals as defined by a transformation T from a region S in the uv-plane onto a region R in the xy-plane
Example 4.1. Let R be the region bounded by the straight lines y = 2x + 3, y = 2x + 1, y = 5 − x and y = 2 − x. Find a transformation T mapping a region S in the uv-plane onto R, where S is a rectangular region, with sides parallel to the u- and v-axes.
The determinant ∂x ∂u ∂x ∂v ∂y ∂u ∂y ∂v
is referred to as the Jacobian of the transformation T and is written using the notation ∂(x, y) ∂(u, v). Theorem 4.1 (Change of Variables in Double Integrals). Suppose that the region S in the uv-plane is mapped onto the region R in the xy-uv-plane by the one-to-one transformation T defined by x = g(u, v) and y = h(u, v), where g and h have continuous first partial derivatives on S. If f is continuous on R and the Jacobian ∂(x, y)
∂(u, v) is nonzero on S, then ¨ R f (x, y)dA = ¨ S f (g(u, v), h(u, v)) ∂(x, y) ∂(u, v) dudv.
Remark. Note that ∂(x, y) ∂(u, v)
stands for the absolute value of the Jacobian. Remark. Sometimes, it is useful to use the formula
∂(u, v) ∂(x, y) = 1 ∂(x, y) ∂(u, v) . Example 4.2. (1) Evaluate ¨ R
(x2+ 2xy) dx dy where R is the region bounded by the lines y = 2x + 3, y = 2x + 1, y = 5 − x, and y = 2 − x.
(2) Evaluate the double integral ¨
R
ex−y
x + ydA, where R is the rectangle bounded by the lines y = x, y = x + 5, y = 2 − x and y = 4 − x.
(3) Find the area of the region in the first quadrant bounded by the curves xy = 1, xy = 4, y = x, and y = 2x.
(4) Evaluate ¨
R
x2
y4 dx dy where R is the region bounded by xy = 2, xy = 4, y
2 = x, and
y2 = 3x.
(5) Evaluate ¨
R
xy dx dy where R is the region bounded by the lines x+3y = −1, x+3y = 3, x − y = 1, and x − y = 2.
(6) Evaluate ¨
R
(x2+y2) dA where R is the region bounded by the lines x+y = 1, x+y = 2, 3x + 4y = 5, and 3x + 4y = 6.
(7) Find the area of the region bounded by the curves y = x2, y = 2x2, x = 3y2, and x = y2. (8) Evaluate
¨
R
(x2+ y2) dA where R is the region in the first quadrant bounded by y = 0, y = x, xy = 1, x2 − y2 = 1.
5. Double Integrals in Polar Coordinates Let
T : x = r cos θ y = r sin θ
be a transformation from rθ-plane to xy-plane. Which transforms regions expressed in polar coordinates to regions in Cartesian coordinates.
Example 5.1 (A Transformation Involving Polar Coordinates). Let R be the region inside the circle x2+ y2 = 9 and outside the circle x2+ y2 = 4 and lying in the first quadrant between the lines y = 0 and y = x. Find a transformation T from a rectangular region S in the rθ-plane to the region R.
Example 5.2. Derive the evaluation formula for polar coordinates (r > 0): ¨ R f (x, y)dA = ¨ S f (r cos θ, r sin θ)rdrdθ.
Theorem 5.1. Suppose that f (r, θ) is continuous on the region R = {(r, θ)|α ≤ θ ≤ β and g1(θ) ≤ r ≤ g2(θ)},
where 0 ≤ g1(θ) ≤ g2(θ) for all θ in [α, β]. Then,
¨ R f (r, θ)dA = ˆ β α ˆ g2(θ) g1(θ) f (r, θ)rdrdθ. Example 5.3. Evaluate ¨ R p
x2+ y2dx dy where R is the disk x2+ y2 ≤ 9.
Example 5.4. Evaluate ¨
R
dx dy
x2+ y2+ 1 where R is the half disk x
Example 5.7. Evaluate ¨
R
(x2+ y2 + 3)dA, where R is the circle of radius 2 centered at the origin.
Example 5.8. Find the volume inside the paraboloid z = 9 − x2 − y2, outside the cylinder
x2+ y2 = 4 and above the xy-plane.
Example 5.9. Evaluate the iterated integral ˆ 1 −1 ˆ √ 1−x2 0 x2(x2+ y2)2dydx.
Example 5.10. Find the volume of the solid bounded by z = 8 − x2− y2 and z = x2+ y2.
Example 5.11. Find the volume cut out of the sphere x2 + y2 + z2 = 4 by the cylinder
Example 5.12. Find the volume of the solid in the first octant bounded by the circular paraboloid z = x2+ y2, the cylinder x2+ y2 = 4 and the coordinate planes.
Example 5.13. Find the volume of the solid under the surface z = x2+ y2 above the xy-plane
and inside the cylinder x2+ y2 = 2y.
Example 5.14. Find the volume V of the solid region T where T is bounded by xy-plane, the cylinder x2+ y2 = x and the cone z =px2+ y2.
Example 5.15. Find the volume V of the solid region T where T is bounded above by the plane z = 1 and below by the cone z =px2+ y2.
Example 5.16. Evaluate ˆ 1 0 ˆ √ 1−x2 0 x2px2+ y2dy dx. 6. Triple Integrals Suppose that f (x, y, z) is continuous on the box Q defined by
Q = {(x, y, z)| a ≤ x ≤ b, c ≤ y ≤ d and r ≤ z ≤ s}. Then, we can write the triple integral over Q as a triple iterated integral:
˚ Q f (x, y, z)dV = ˆ s r ˆ d c ˆ b a f (x, y, z)dxdydz.
Example 6.1. Evaluate the triple integral ˚
Q
2xeysin zdV , where Q is the rectangle defined by
Q = {(x, y, z)| 1 ≤ x ≤ 2, 0 ≤ y ≤ 1 and 0 ≤ z ≤ π}.
In general, if F (x, y, z) is a function defined on a closed bounded region Q in space, we evaluate the triple integral by applying a three dimensional version of iterated integration.
For instance, if the region Q can be written in the form
where R is some region in the xy-plane and where g1(x, y) ≤ g2(x, y) for all (x, y) in R, then ˚ Q f (x, y, z)dV = ¨ R ˆ g2(x,y) g1(x,y) f (x, y, z)dzdA. Example 6.2. Evaluate ˚ Q
6xydV , where Q is the tetrahedron bounded by the planes x = 0, y = 0, z = 0 and 2x + y + z = 4.
Example 6.3. Evaluate ˚
Q
6xydV , where Q is the tetrahedron bounded by the planes x = 0, y = 0, z = 0 and 2x + y + z = 4 as in the previous example, but this time, integrate first with respect to x. Example 6.4. Evaluate ˆ 4 0 ˆ 4 x ˆ y 0 6 1 + 48z − z3dzdydx.
Recall that for double integrals, we had mentioned that ¨
R
1dA gives the area of the region R. Similarly for solid region in space
V = ˚
Q
1dV
where V is the volume of the solid Q.
Example 6.5. Find the volume of the solid bounded by the graphs of z = 4 − y2, x + z = 4,
Example 6.6. Find the volume of the solid region bounded by the planes x = 0, x = 3, z = 1 and z = y2.
Steps to evaluate a triple by integration over a solid region Q first with respect to z, then with respect to y, and finally with respect to x.
• Step 1. Sketch: Sketch the region Q along with its “shadow” R (vertical projection) in the xy-plane. Label the upper and lower bounding surfaces of Q and the upper and lower bounding curves of R.
• Step 2. Find the z- limits of integration: Draw a line M passing through a typical point (x, y) in R parallel to the z- axis. As z increases, M enters Q at z = f1(x, y) and
leaves at z = f2(x, y). These are the z- limits of integration.
• Step 3. Find the y- limits of integration: Draw a line L through (x, y) parallel to the y- axis. As y increases, L enters R at y = g1(x) and leaves at y = g2(x). These are
the y- limits of integration.
• Step 4. Find the x- limits of integration: Choose x- limits that include all lines through R parallel to the y- axis. (x = a, x = b).
Then the integral is ˚ Q F (x, y, z)dV = ˆ x=b x=a ˆ y=g2(x) y=g1(x) ˆ z=f2(x,y) z=f1(x,y) F (x, y, z) dz dy dx.
Follow a similar procedure if you change the order of integration.
Change of Variables in Triple Integrals. For a transformation T from a region S of uvw-space onto a region R in xyz-uvw-space, defined by x = g(u, v, w), y = h(u, v, w) and z = l(u, v, w), the Jacobian of the transformation is the determinant ∂(x, y, z)
∂(u, v, w) defined by ∂(x, y, z) ∂(u, v, w) = ∂x ∂u ∂x ∂v ∂x ∂w ∂y ∂u ∂y ∂v ∂z ∂w ∂z ∂u ∂z ∂v ∂z ∂w
Theorem 6.1. Suppose that the region S in uvw-space is mapped onto the region R in xyz-space by the one-to-one transformation T defined by x = g(u, v, w), y = h(u, v, w) and z = l(u, v, w), where g, h and l have continuous first partial derivatives in S. If f is continuous in R and the
7. Cylindrical Coordinates
Example 7.1. Derive the evaluation formula for triple integrals in spherical coordinates: ˚ R f (x, y, z)dV = ˚ S f (r cos θ, r sin θ, z)r dz dr dθ. Example 7.2. Evaluate ˚ Q
ex2+y2dV , where Q is the solid bounded by the cylinder x2+y2 = 9,
the xy-plane and the plane z = 5.
Example 7.3. Write ˚
Q
f (r, θ, z)dV as a triple iterated integral in cylindrical coordinates if
Example 7.4. Evaluate the triple iterated integral ˆ 1 −1 ˆ √ 1−x2 −√1−x2 ˆ 2−x2−y2 x2+y2 (x2+ y2)32dzdydz.
Example 7.5. Use a triple integral to find the volume of the solid Q bounded by the graph of y = 4 − x2 − z2 and the xz-plane.
Example 7.6. Find the volume V of D where D is the solid region bounded by z = x2 + y2
and z =px2+ y2. Example 7.7. Evaluate ˆ 2 0 ˆ √ 2x−x2 0 ˆ 3 0
zpx2+ y2dz dy dx using cylindrical coordinates.
Example 7.8. Evaluate ˚
D
(x2+ y2+ 1) dz dy dx where D is the solid region bounded by the cylinder x2+ y2 = 2y and the panes z = 0 and z = 2.
Example 7.9. Find the volume V of the solid region bounded by the plane x + z = 1 and the paraboloid z = 1 + x2+ y2.
Example 7.10. Evaluate ˚
D
zy dV where D is the solid region bounded above by the plane z = 1 and below by the cone z =px2+ y2.
Example 7.11. Evaluate ˚
D
xy dV where D is the solid region in the first octant bounded above by the the hemisphere z = p4 − x2− y2 and on the sides and the bottom by the
Example 7.14. Evaluate ˆ 1 0 ˆ √ 1−x2 −√1−x2 ˆ 1 x2+y2 x dz dy dx. 8. Spherical Coordinates
Example 8.1. Derive the evaluation formula for triple integrals in spherical coordinates: ˚
R
f (x, y, z)dV = ˚
S
f (ρ sin φ cos θ, ρ sin φ sin θ, ρ cos φ)ρ2sin φ dρ dφ dθ.
Example 8.2. Find the volume of the sphere of radius a. Example 8.3. Evaluate the triple integral˝
Q
cos(x2+ y2+ z2)32dV , where Q is the unit ball:
x2+ y2+ z2 ≤ 1.
Example 8.4. Find the volume lying inside the sphere x2+ y2+ z2 = 2z and inside the cone
z2 = x2+ y2.
Example 8.5. Evaluate the triple iterated integral ˆ 2 −2 ˆ √ 4−x2 0 ˆ √4−x2−y2 0 (x2+y2+z2)dzdydx.
Example 8.6. Evaluate ˚
D
p
x2+ y2+ z2dx dy dz, where D is the ball x2+ y2+ z2 ≤ a2.
Example 8.7. Evaluate ˚
D
xz dx dy dz, where D is the part of the spherical shell 1 ≤ x2+
