16.2-Iterated Integrals
Writing a double integral as an iterated integral: if R is a rectangle a<=x<=b, c<=y<=d, and f is a
continuous function on R, then the integral of f over R exists and is equal to the iterated integral. Which
is: int(c,d) int (a,b) f(x,y)dxdy
Q: Sketch the region of integration for the iterated integral int(0,6) int(x/3, 2) x*sqrt(y^3+1)dydx
A: y is limited by x/3 and 2 and x is limited by 0 and 6. Graph the equation then color in within those
regions
Q: Evaluate (0,6) int(x/3, 2) x*sqrt(y^3+1)dydx using the region sketched.
A: Step one is reverse the order of integration. From x=0 to x=3y. int(0,2) int(0,3y) x*sqrt(y^3+1) dxdy
then do the integrals out.
16.3-Triple Integrals
Q: A cube C has sides of length 4cm. If one corner is on the origin and the adjacent corners are on the
positive x,y, and z axes, then density at the point (x, y, z) is symbol(x, y, z)=1+xyz gm/cm^3. Find the
mass of the cube
A: mass of small piece=density * volume (vol= symbol(x, y, z) DELTA V). M=int (0,4) int (0,4) int
(0,4)(1+xyz)dxdydz. Do it out.
Q: setup an iterated integral to compute the mass of the solid cone bounded by z=sqrt(x^2 + y^2) and
z=3, if the denisty is given by symbol(x, y, z)=z.
A:int(sqrt(x^2 to y^2), 3) zdz. Int(-sqrt(9-x^2), sqrt(9-x^2)) int(sqrt(x^2 to y^2), 3) zdz. Then int(-3, 3)Int(sqrt(9-x^2), sqrt(9-x^2)) int(sqrt(x^2 to y^2), 3) zdzdydx then do it out.
16.4-double integrals in polar coordinates
X=rcos(theta)
Y=rsin(theta)
(x^2 + y^2)=r^2
dA=rdrd(theta)
Q: compute the integral of f(x,y) = 1/ (x^2 + y^2)^(3/2) over the region R.
A: Region R = 1<=r<=2, 0<=theta<=pi/4. In polar r=sqrt(x^2 + y^2) so f(x,y)= 1/ (x^2 + y^2)^(3/2) =
1/(r^2)^(3/2) = 1/r^3. Then int(0, pi/4)int(1, 2) of 1/r^3 rdrd(theta). Then do it out
16.5-integrals in cylindrical and spherical coordinates
Relationship between Cartesian and cylindrical coordinates: 0<=r<infinity, 0<=theta<=2pi, -inf<z<inf.
X=rcos(theta) y=rsin(theta) z=z and x^2 + y^2 = r^2
dV=rdrd(theta)dz
Relationship between Cartesian and spherical coordinates: 0<=p<inf, 0<=phi<=pi, 0<=theta<=2pi.
X=psin(phi)cos(theta) y=psin(phi)sin(theta) z=pcos(phi) p^2=x^2+y^2+z^2
Q: A tank in the shape of a hemisphere has radius a; its base is its plane face. Fine the volume, V, of the
water in the tank as a function of h, the depth of water.
A: in cart a^2=x^2+y^2+z^2, in cylin r^2=x^2+y^2 so it becomes r^2+z^2=a^2. Thus, we let r go from 0
to sqrt(a^2-z^2), we let (theta) go from 0 to 2pi and we let z go from 0 to h giving: int(0, 2pi)int(0,
h)int(0, sqrt(a^2-z^2))rdrdzd(theta). Then do it out.
Q: use spherical coordinates to derive the formula for the volume of a solid sphere of radius a
A: in spherical, a sphere of radius a is 0<=p<=a, 0<=theta<=2pi and 0<=phi<=pi. Volume= int(0, 2pi)int(0,
pi)int(0, a) p^2 sin(phi)dpd(phi)d(theta)== int(0, 2pi)int(0,
pi)(1/3)a^3sin(phi)d(phi)d(theta)==(1/3)a^3int(0, 2pi)-cos(phi) from (0, pi)d(theta)==(2/3)a^3int (0,
2pi)d(theta)==(4pia^3)/3
17.1-parameterized curves
Parametric equations of a line through point (x0,y0,z0) and parallel to the vector ai+bj+ck are x=x0+at
y=y0+bt z=z0+ct
Parametric equation of a line in vector form: the line through the point with position vector
r0=x0i+y0j+z0k in the direction of vector v=ai+bj+ck has parametric equation r(t)=r0+tv
Q: line in the direction of the vector 2i+3j+4k and through point (1, 5, 7)
A: x=1+2t y=5+3t z=7+4t
Q: the line through points (1,2,-1) and (3,3,4)
A: P=(1,2,-1) and Q=(3,3,4) so PQ=(3-1)I + (3-2)j + (4-(-1))k=2i+j+5k. P-> x=1+2t y=2+t z=-1+5t AND Q->
x=3+2t y=3+t z=4+5t
Q; the circle with radius ½ centered @ pt (-1,2)
A: r1(t)=costi +sintj; pt (-1,2) has position vector –i+2j. position vector r(t)=1/2r1+r0. So r(t)=1/2(costi +
sintj)+(-i+2j)=(-1+1/2cost)i+(2_1/2sint)j
Q: Two particles move through space, with equations r1(t)=ti+(1+2t)j+(3-2t)k and r2=(-2-2t)i+(12t)j+(1+t)k. do the particles ever collide? Do their paths cross?
A: to test if they collide, r1(t)=r2(t). t=-2-2t 1+2t=1-2t 3-2t=1+t which gives t=-2/3, 0, and 2/3 so
since no common solution they don’t collide. To see if they cross paths we do t1=-2-2t2 1+2t1=1-2t2
3-2t1=1+t2 and get t1=2, t2=-2 since these two satisfy third equation the paths cross.
17.2
Velocity vector=f`(t)i+g`(t)j+h`(t)k
Accel vector=same as velocity but second derive
Uniform circle: r(t)=Rcos(wt)i+Rsin(wt)j
Straight line: r(t)=r0+f(t)v0