# CSCI 2170 Study Guide - Fall 2019, Comprehensive Final Exam Notes - Euclidean Vector, Wireless Access Point, Integer

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CSCI 2170

Math 241 Chapter 14 Dr. Justin O. Wyss-Gallifent

§14.1 Double Integrals

1. These can be deﬁned via a Riemann Sum method like in Calculus I but the net result is: We can

deﬁne the double integral of f(x, y)over R, denoted RRRf(x, y)dA to be the signed volume under

the graph of f(x, y) within the region R. The question is how to evaluate these things. First...

2. Defn: An iterated integral is a nested integral. An inner integral may have limits of integration

which include variables further out. We evaluate these by working from the inside out, making

sure we integrate with respect to the correct variable each time.

3. Now then, onto evaluation of RRRf(x, y)dA.

(a) Ris vertically simple if Rmay be described as between the two functions y=bot(x) and

y=top(x) on the interval a≤x≤b. In this case RR

R

f(x, y)dA =Rb

aRtop

bot f(x, y)dy dx.

(b) Ris horizontally simple if Rmay be described as between the two functions y=lef t(x) and

y=right(x) on the interval c≤y≤d. In this case RR

R

f(x, y)dA =Rd

cRright

lef t f(x, y)dx dy.

Consider in both cases that if either the top, bottom, left or right function ever changes then you

will need more than one integral.

4. We can reparametrize (HS to VS or VS to HS) to do an impossible integral like R1

0R1

xe(y2)dy dx.

§14.2 Double Integrals in Polar Coordinates

1. Reminder about how polar coordinates work. Shapes we’ll see a lot include things like r= 2,

r= 3 cos θ,r= 2 sin θ,r= 1 + cos θas well as vertical and horizontal lines which need to be

converted. Don’t forget x=rcos θ,y=rsin θand x2+y2=r2.

2. We describe a region in polar coordinates from the point of view of a person who lives at the

origin. There is a near function r=near(θ) and a far function r=far(θ) between two angles

α≤θ≤β. In this case RR

R

f(x, y)dA =Rβ

αRfar

near f(rcos θ, r sin θ)r dr dθ. We’ll see later where

that extra rcomes from. It might help to remember it’s the “Jacobian r”.

3. We can reparametrize (to polar) to do an impossible integral like R1

−1R√1−x2

0sin(x2+y2)dy dx.

§14.4 Triple Integrals

1. Finding a volume analogy is tricky. Instead suppose Dis a solid object in space and at any point

f(x, y, z) is the density around that point. Then we can deﬁne RRR

D

f(x, y, z)dV as the mass of

D. The question is how to evaluate which all depends upon how to best describe D.

2. We have the following:

(a) If Dis the solid between the graphs of z=low(x, y) and z=high(x, y) above the region R

in the xy-plane and if Ris VS then RRR

D

f(x, y, z)dV =Rb

aRtop

bot Rhigh

low f(x, y, z)dz dy dx.

(b) If Dis the solid between the graphs of z=low(x, y) and z=high(x, y) above the region R

in the xy-plane and if Ris HS then RRR

D

f(x, y, z)dV =Rd

cRright

lef t Rhigh

low f(x, y, z)dz dx dy.

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§14.5 Triple Integrals in Cylindrical Coordinates

1. Cylindrical coordinates are just polar coordinates plus z. The thing to watch out for is how

equations change. For example:

(a) r= 2 is a cylinder, as are r= 3 cos θand r= 2 sin θ.

(b) The sphere x2+y2+z2= 9 becomes r2+z2= 9.

(c) The cone z=px2+y2becomes z=r.

(d) The plane x= 2 becomes rcos θ= 2 or r= 2 sec θ.

2. If Dis the solid between the graphs of z=low(x, y) and z=high(x, y) above the region Rin the

xy-plane and if Ris polar then we have to convert low and high to polar functions z=low(r, θ)

and z=high(r, θ) in terms of rand/or θand then

RRR

D

f(x, y, z)dV =Rβ

αRfar

near Rhigh

low f(rcos θ, r sin θ, z)r dz dθ dr.

§14.6 Triple Integrals in Spherical Coordinates

1. Describe how spherical coordinates work and make sure to mention the conversions:

(a) x=ρsin φcos θ

(b) y=ρsin φsin θ

(c) z=ρcos φ

(d) x2+y2+z2=ρ2

(e) x2+y2=ρ2sin2φ

2. Equations can change here. For example:

(a) ρ= 2 is a sphere.

(b) The cylinder x2+y2= 4 becomes ρ= 2 csc φ.

(c) φ=π

4is a cone.

(d) The plane z= 3 becomes ρ= 3 sec φ.

3. To describe a solid in spherical we take a range α≤θ≤βand γ≤φ≤δFrom the point of view

of a person at the origin this describes a “window” looking out. In that window we have a near

function ρ=near(φ, θ) and a far function ρ=far(φ, θ).

4. If Dis described this way then

RRR

D

f(x, y, z)dV =Rβ

αRδ

γRfar

near f(ρsin φcos θ, ρ sin φsin θ, ρ cos φ)ρ2sin φ dρ dφ dθ.

Don’t forget that ρ2sin φ. It’s the “Jacobian” again.

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