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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