Multiple Integrals Christopher Croke University of Pennsylvania Math 115 Christopher Croke Calculus 115
Multiple Integrals
Christopher Croke
University of Pennsylvania
Math 115
Christopher Croke Calculus 115
Integrals.
The geometry of integrals of functions of one variable∫ ba f (x)dx :
For two variables∫ ∫
R f (x , y)dA:
Volume under x , y plane counts negative.
Christopher Croke Calculus 115
Integrals.
The geometry of integrals of functions of one variable∫ ba f (x)dx :
For two variables∫ ∫
R f (x , y)dA:
Volume under x , y plane counts negative.
Christopher Croke Calculus 115
How to calculate
Use iterated integrals.
∫ b
a
[ ∫ h(x)
g(x)f (x , y)dy
]dx
where a and b are constants and g(x) and h(x) are functions of xonly.How to compute?First find an antiderivative F (x , y) thinking of x as a constant,that is:
∂F
∂y(x , y) = f (x , y).
Then compute: ∫ b
a
[F (x , h(x)) − F (x , g(x))
]dx .
(which is just an integral of a function of one variable.)
Christopher Croke Calculus 115
How to calculate
Use iterated integrals.∫ b
a
[ ∫ h(x)
g(x)f (x , y)dy
]dx
where a and b are constants and g(x) and h(x) are functions of xonly.
How to compute?First find an antiderivative F (x , y) thinking of x as a constant,that is:
∂F
∂y(x , y) = f (x , y).
Then compute: ∫ b
a
[F (x , h(x)) − F (x , g(x))
]dx .
(which is just an integral of a function of one variable.)
Christopher Croke Calculus 115
How to calculate
Use iterated integrals.∫ b
a
[ ∫ h(x)
g(x)f (x , y)dy
]dx
where a and b are constants and g(x) and h(x) are functions of xonly.How to compute?
First find an antiderivative F (x , y) thinking of x as a constant,that is:
∂F
∂y(x , y) = f (x , y).
Then compute: ∫ b
a
[F (x , h(x)) − F (x , g(x))
]dx .
(which is just an integral of a function of one variable.)
Christopher Croke Calculus 115
How to calculate
Use iterated integrals.∫ b
a
[ ∫ h(x)
g(x)f (x , y)dy
]dx
where a and b are constants and g(x) and h(x) are functions of xonly.How to compute?First find an antiderivative F (x , y) thinking of x as a constant,that is:
∂F
∂y(x , y) = f (x , y).
Then compute: ∫ b
a
[F (x , h(x)) − F (x , g(x))
]dx .
(which is just an integral of a function of one variable.)
Christopher Croke Calculus 115
How to calculate
Use iterated integrals.∫ b
a
[ ∫ h(x)
g(x)f (x , y)dy
]dx
where a and b are constants and g(x) and h(x) are functions of xonly.How to compute?First find an antiderivative F (x , y) thinking of x as a constant,that is:
∂F
∂y(x , y) = f (x , y).
Then compute: ∫ b
a
[F (x , h(x)) − F (x , g(x))
]dx .
(which is just an integral of a function of one variable.)Christopher Croke Calculus 115
Problem: Compute: ∫ 2
0
[ ∫ 4
3(x2y)dy
]dx .
(You don’t need the [, ]’s)
Problem: Compute: ∫ 1
−1
∫ 1+x
x2xydydx .
Christopher Croke Calculus 115
Problem: Compute: ∫ 2
0
[ ∫ 4
3(x2y)dy
]dx .
(You don’t need the [, ]’s)
Problem: Compute: ∫ 1
−1
∫ 1+x
x2xydydx .
Christopher Croke Calculus 115
Problem: Compute: ∫ 2
0
[ ∫ 4
3(x2y)dy
]dx .
(You don’t need the [, ]’s)
Problem: Compute: ∫ 1
−1
∫ 1+x
x2xydydx .
Christopher Croke Calculus 115
How does ∫ b
a
[ ∫ g2(x)
g1(x)f (x , y)dy
]dx
relate to ∫ ∫Rf (x , y)dA?
It is the answer when R is the region:
Christopher Croke Calculus 115
How does ∫ b
a
[ ∫ g2(x)
g1(x)f (x , y)dy
]dx
relate to ∫ ∫Rf (x , y)dA?
It is the answer when R is the region:
Christopher Croke Calculus 115
Problem: Calculate the volume of the solid bounded above byf (x , y) = x2 over the rectangle 0 ≤ x ≤ 1, 2 ≤ y ≤ 3.
Problem: Calculate ∫ ∫R
(x + y)dydx
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.Problem *: Calculate
∫ ∫R y dA where R is the region in the
triangle with vertices (−1, 0),(1, 0), and (0, 1).
Christopher Croke Calculus 115
Problem: Calculate the volume of the solid bounded above byf (x , y) = x2 over the rectangle 0 ≤ x ≤ 1, 2 ≤ y ≤ 3.
Problem: Calculate ∫ ∫R
(x + y)dydx
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.Problem *: Calculate
∫ ∫R y dA where R is the region in the
triangle with vertices (−1, 0),(1, 0), and (0, 1).
Christopher Croke Calculus 115
Problem: Calculate the volume of the solid bounded above byf (x , y) = x2 over the rectangle 0 ≤ x ≤ 1, 2 ≤ y ≤ 3.
Problem: Calculate ∫ ∫R
(x + y)dydx
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.
Problem *: Calculate∫ ∫
R y dA where R is the region in thetriangle with vertices (−1, 0),(1, 0), and (0, 1).
Christopher Croke Calculus 115
Problem: Calculate the volume of the solid bounded above byf (x , y) = x2 over the rectangle 0 ≤ x ≤ 1, 2 ≤ y ≤ 3.
Problem: Calculate ∫ ∫R
(x + y)dydx
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.Problem *: Calculate
∫ ∫R y dA where R is the region in the
triangle with vertices (−1, 0),(1, 0), and (0, 1).
Christopher Croke Calculus 115
Often better to integrate w.r.t. x first. Fubini’s Theorem says weget the same answer.
∫ d
c
[ ∫ h2(y)
h1(y)f (x , y)dx
]dy
corresponds to the region
Christopher Croke Calculus 115
Often better to integrate w.r.t. x first. Fubini’s Theorem says weget the same answer.∫ d
c
[ ∫ h2(y)
h1(y)f (x , y)dx
]dy
corresponds to the region
Christopher Croke Calculus 115
Easy example:∫ 1
0
∫ 2
0(x + y)dydx =
∫ 2
0
∫ 1
0(x + y)dxdy .
Now we will do Problem* this way.
Sometimes you *have* to do it this way!:Problem: Compute: ∫ 1
0
∫ 1
xey
2dydx .
There is no good rule to tell you which way to go. Often the shapeof the region gives a hint.
Christopher Croke Calculus 115
Easy example:∫ 1
0
∫ 2
0(x + y)dydx =
∫ 2
0
∫ 1
0(x + y)dxdy .
Now we will do Problem* this way.
Sometimes you *have* to do it this way!:Problem: Compute: ∫ 1
0
∫ 1
xey
2dydx .
There is no good rule to tell you which way to go. Often the shapeof the region gives a hint.
Christopher Croke Calculus 115
Easy example:∫ 1
0
∫ 2
0(x + y)dydx =
∫ 2
0
∫ 1
0(x + y)dxdy .
Now we will do Problem* this way.
Sometimes you *have* to do it this way!:Problem: Compute: ∫ 1
0
∫ 1
xey
2dydx .
There is no good rule to tell you which way to go. Often the shapeof the region gives a hint.
Christopher Croke Calculus 115
Easy example:∫ 1
0
∫ 2
0(x + y)dydx =
∫ 2
0
∫ 1
0(x + y)dxdy .
Now we will do Problem* this way.
Sometimes you *have* to do it this way!:Problem: Compute: ∫ 1
0
∫ 1
xey
2dydx .
There is no good rule to tell you which way to go. Often the shapeof the region gives a hint.
Christopher Croke Calculus 115
Problem: Find the volume of the solid in the first octant boundedby the coordinate planes, the cylinder x2 + y2 = 1, and the planez − 2y = 0.
Sometimes can integrate over unbounded regions:Problem: Compute: ∫ ∞
1
∫ 1x
0y3xdydx .
Christopher Croke Calculus 115
Problem: Find the volume of the solid in the first octant boundedby the coordinate planes, the cylinder x2 + y2 = 1, and the planez − 2y = 0.
Sometimes can integrate over unbounded regions:
Problem: Compute: ∫ ∞1
∫ 1x
0y3xdydx .
Christopher Croke Calculus 115
Problem: Find the volume of the solid in the first octant boundedby the coordinate planes, the cylinder x2 + y2 = 1, and the planez − 2y = 0.
Sometimes can integrate over unbounded regions:Problem: Compute: ∫ ∞
1
∫ 1x
0y3xdydx .
Christopher Croke Calculus 115
Problem: Find the volume of the solid in the first octant boundedby the coordinate planes, the cylinder x2 + y2 = 1, and the planez − 2y = 0.
Sometimes can integrate over unbounded regions:Problem: Compute: ∫ ∞
1
∫ 1x
0y3xdydx .
Christopher Croke Calculus 115
Riemann sums
Break your rectangular region into smaller rectangles Rk of areaA(Rk).
Choose a point (xk , yk) in each Rk . Then the Riemannsum that approximates the integral
∫ ∫R f (x , y)dA is:
Σk f (xk , yk)A(Rk).
Christopher Croke Calculus 115
Riemann sums
Break your rectangular region into smaller rectangles Rk of areaA(Rk). Choose a point (xk , yk) in each Rk .
Then the Riemannsum that approximates the integral
∫ ∫R f (x , y)dA is:
Σk f (xk , yk)A(Rk).
Christopher Croke Calculus 115
Riemann sums
Break your rectangular region into smaller rectangles Rk of areaA(Rk). Choose a point (xk , yk) in each Rk . Then the Riemannsum that approximates the integral
∫ ∫R f (x , y)dA is:
Σk f (xk , yk)A(Rk).
Christopher Croke Calculus 115
Riemann sums
Break your rectangular region into smaller rectangles Rk of areaA(Rk). Choose a point (xk , yk) in each Rk . Then the Riemannsum that approximates the integral
∫ ∫R f (x , y)dA is:
Σk f (xk , yk)A(Rk).
Christopher Croke Calculus 115
Riemann sums
Break your rectangular region into smaller rectangles Rk of areaA(Rk). Choose a point (xk , yk) in each Rk . Then the Riemannsum that approximates the integral
∫ ∫R f (x , y)dA is:
Σk f (xk , yk)A(Rk).
Christopher Croke Calculus 115
There is a theorem which says that the limit as the size of therectangles Rk goes to 0 then the Riemann sum approaches theintegral.
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
Christopher Croke Calculus 115
There is a theorem which says that the limit as the size of therectangles Rk goes to 0 then the Riemann sum approaches theintegral. 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
Christopher Croke Calculus 115
There is a theorem which says that the limit as the size of therectangles Rk goes to 0 then the Riemann sum approaches theintegral. 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
Christopher Croke Calculus 115
There is a theorem which says that the limit as the size of therectangles Rk goes to 0 then the Riemann sum approaches theintegral. 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
Christopher Croke Calculus 115