Download Multiple Integrals - Penn Math

Christopher Croke
University of Pennsylvania
Multiple Integrals
Math 115
Calculus 115
Integrals.
The geometry of integrals of functions of one variable
Calculus 115
Rb
a
f (x)dx:
Integrals.
The geometry of integrals of functions of one variable
For two variables
RR
R
Rb
a
f (x)dx:
f (x, y )dA:
Volume under x, y plane counts negative.
Calculus 115
How to calculate
Use iterated integrals.
Calculus 115
How to calculate
Use iterated integrals.
Z b hZ
a
h(x)
i
f (x, y )dy dx
g (x)
where a and b are constants and g (x) and h(x) are functions of x
only.
Calculus 115
How to calculate
Use iterated integrals.
Z b hZ
a
h(x)
i
f (x, y )dy dx
g (x)
where a and b are constants and g (x) and h(x) are functions of x
only.
How to compute?
Calculus 115
How to calculate
Use iterated integrals.
Z b hZ
a
h(x)
i
f (x, y )dy dx
g (x)
where a and b are constants and g (x) and h(x) are functions of x
only.
How to compute?
First find an antiderivative F (x, y ) thinking of x as a constant,
that is:
∂F
(x, y ) = f (x, y ).
∂y
Calculus 115
How to calculate
Use iterated integrals.
Z b hZ
a
h(x)
i
f (x, y )dy dx
g (x)
where a and b are constants and g (x) and h(x) are functions of x
only.
How to compute?
First find an antiderivative F (x, y ) thinking of x as a constant,
that is:
∂F
(x, y ) = f (x, y ).
∂y
Then compute:
Z
b
h
i
F (x, h(x)) − F (x, g (x)) dx.
a
(which is just an integral of a function of one variable.)
Calculus 115
Problem: Compute:
Z
0
2hZ 4
i
(x 2 y )dy dx.
3
Calculus 115
Problem: Compute:
Z
2hZ 4
0
i
(x 2 y )dy dx.
3
(You don’t need the [, ]’s)
Calculus 115
Problem: Compute:
2hZ 4
Z
0
i
(x 2 y )dy dx.
3
(You don’t need the [, ]’s)
Problem: Compute:
Z
1
Z
1+x
xydydx.
−1
x2
Calculus 115
How does
Z
a
relate to
b
hZ
g2 (x)
i
f (x, y )dy dx
g1 (x)
Z Z
f (x, y )dA?
R
Calculus 115
How does
Z
a
relate to
b
hZ
g2 (x)
i
f (x, y )dy dx
g1 (x)
Z Z
f (x, y )dA?
R
It is the answer when R is the region:
Calculus 115
Problem: Calculate the volume of the solid bounded above by
f (x, y ) = x 2 over the rectangle 0 ≤ x ≤ 1, 2 ≤ y ≤ 3.
Calculus 115
Problem: Calculate the volume of the solid bounded above by
f (x, y ) = x 2 over the rectangle 0 ≤ x ≤ 1, 2 ≤ y ≤ 3.
Problem: Calculate
Z Z
(x + y )dydx
R
where R is the region in the triangle with vertices (0, 0), (2, 0) and
(0, 1).
Calculus 115
Problem: Calculate the volume of the solid bounded above by
f (x, y ) = x 2 over the rectangle 0 ≤ x ≤ 1, 2 ≤ y ≤ 3.
Problem: Calculate
Z Z
(x + y )dydx
R
where R is the region in the triangle with vertices (0, 0), (2, 0) and
(0, 1).
Sometimes need to split the region into two (or more) pieces.
Calculus 115
Problem: Calculate the volume of the solid bounded above by
f (x, y ) = x 2 over the rectangle 0 ≤ x ≤ 1, 2 ≤ y ≤ 3.
Problem: Calculate
Z Z
(x + y )dydx
R
where R is the region in the triangle with vertices (0, 0), (2, 0) and
(0, 1).
Sometimes need to splitR the
R region into two (or more) pieces.
Problem *: Calculate
R y dA where R is the region in the
triangle with vertices (−1, 0),(1, 0), and (0, 1).
Calculus 115
Often better to integrate w.r.t. x first. Fubini’s Theorem says we
get the same answer.
Calculus 115
Often better to integrate w.r.t. x first. Fubini’s Theorem says we
get the same answer.
Z
d
hZ
c
h2 (y )
i
f (x, y )dx dy
h1 (y )
corresponds to the region
Calculus 115
Easy example:
Z 1Z
2
Z
2Z 1
(x + y )dydx =
0
0
(x + y )dxdy .
0
0
Calculus 115
Easy example:
Z 1Z
2
Z
2Z 1
(x + y )dydx =
0
0
(x + y )dxdy .
0
0
Now we will do Problem* this way.
Calculus 115
Easy example:
Z 1Z
2
Z
2Z 1
(x + y )dydx =
0
0
(x + y )dxdy .
0
0
Now we will do Problem* this way.
Sometimes you *have* to do it this way!:
Problem: Compute:
Z 1Z 1
2
e y dydx.
0
x
Calculus 115
Easy example:
Z 1Z
2
Z
2Z 1
(x + y )dydx =
0
0
(x + y )dxdy .
0
0
Now we will do Problem* this way.
Sometimes you *have* to do it this way!:
Problem: Compute:
Z 1Z 1
2
e y dydx.
0
x
There is no good rule to tell you which way to go. Often the shape
of the region gives a hint.
Calculus 115
Problem: Find the volume of the solid in the first octant bounded
by the coordinate planes, the cylinder x 2 + y 2 = 1, and the plane
z − 2y = 0.
Calculus 115
Problem: Find the volume of the solid in the first octant bounded
by the coordinate planes, the cylinder x 2 + y 2 = 1, and the plane
z − 2y = 0.
Sometimes can integrate over unbounded regions:
Calculus 115
Problem: Find the volume of the solid in the first octant bounded
by the coordinate planes, the cylinder x 2 + y 2 = 1, and the plane
z − 2y = 0.
Sometimes can integrate over unbounded regions:
Problem: Compute:
Z
1
∞Z
1
x
y 3 xdydx.
0
Calculus 115
Problem: Find the volume of the solid in the first octant bounded
by the coordinate planes, the cylinder x 2 + y 2 = 1, and the plane
z − 2y = 0.
Sometimes can integrate over unbounded regions:
Problem: Compute:
Z
1
∞Z
1
x
y 3 xdydx.
0
Calculus 115
Riemann sums
Break your rectangular region into smaller rectangles Rk of area
A(Rk ).
Calculus 115
Riemann sums
Break your rectangular region into smaller rectangles Rk of area
A(Rk ). Choose a point (xk , yk ) in each Rk .
Calculus 115
Riemann sums
Break your rectangular region into smaller rectangles Rk of area
A(Rk ). Choose a point (xk , yk ) in each
R R Rk . Then the Riemann
sum that approximates the integral
R f (x, y )dA is:
Σk f (xk , yk )A(Rk ).
Calculus 115
Riemann sums
Break your rectangular region into smaller rectangles Rk of area
A(Rk ). Choose a point (xk , yk ) in each
R R Rk . Then the Riemann
sum that approximates the integral
R f (x, y )dA is:
Σk f (xk , yk )A(Rk ).
Calculus 115
Riemann sums
Break your rectangular region into smaller rectangles Rk of area
A(Rk ). Choose a point (xk , yk ) in each
R R Rk . Then the Riemann
sum that approximates the integral
R f (x, y )dA is:
Σk f (xk , yk )A(Rk ).
Calculus 115
There is a theorem which says that the limit as the size of the
rectangles Rk goes to 0 then the Riemann sum approaches the
integral.
Calculus 115
There is a theorem which says that the limit as the size of the
rectangles Rk goes to 0 then the Riemann sum approaches the
integral. You can see from the picture why Fubini’s theorem works.
Calculus 115
There is a theorem which says that the limit as the size of the
rectangles Rk goes to 0 then the Riemann sum approaches the
integral. You can see from the picture why Fubini’s theorem works.
You can do the same sort of thing if the region is not a rectangle.
Calculus 115
There is a theorem which says that the limit as the size of the
rectangles Rk goes to 0 then the Riemann sum approaches the
integral. You can see from the picture why Fubini’s theorem works.
You can do the same sort of thing if the region is not a rectangle.
END OF MATERIAL FOR THE MIDTERM
Calculus 115