christopher croke - university of pennsylvania

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.


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.)


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 .



How to calculate


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 .



How to calculate


a

[ ∫ h(x)

g(x)f (x , y)dy

]dx


∂F

∂y(x , y) = f (x , y).

Then compute: ∫ b

a

[F (x , h(x)) − F (x , g(x))

]dx .



How to calculate


a

[ ∫ h(x)

g(x)f (x , y)dy

]dx


∂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 .


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:


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).




(x + y)dydx


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).




(x + y)dydx


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).


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


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


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.


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 .



Sometimes can integrate over unbounded regions:

Problem: Compute: ∫ ∞1

∫ 1x

0y3xdydx .



Sometimes can integrate over unbounded regions:Problem: Compute: ∫ ∞

1

∫ 1x

0y3xdydx .


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).


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




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




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


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.



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.



christopher croke - university of pennsylvania

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