course mth277
Find surface area of the porti0n of the sphere x^2+y^2+z^2=4 that lies inside the cylinder x^2+y^2=2y
I noted that the sphere radius is 2, I also broke down the cylinder to the form x^2 + (y-1)^2 =1
giving the form y= 'sqrt (1-x^2)+1
z=(f(x,y))= sqrt (4-x^2-y^2)
fx= -x/sqrt (4-x^2-y^2)
fy= -y/sqrt (4-x^2-y^2)
In the picture it shows that the cylinder runs from 0 to 1 then it shows the sphere with radius 2 as I
previously stated
so I wrote 0
and 0
those are the limits on the double integral in the form sqrt((fx)^2+(fy)^2 +1) It gets pretty messy
so I haven't done the integration all the way out, mostly because I just want to make it set up right
and integrating these functions aren't too difficult.
The circle x^2 + (y-1)^2 =1 is centered at (0, 1) and has radius 1.
If you set this up with vertical cross-sections (i.e., partition the x interval), x goes from -1 to 1. If you've sketched the circle you see that the entire circle lies above the x axis, so that for any x (except -1 and 1) the y interval is bounded below and above by the circle.
Solve the equation for y:
(y - 1)^2 = 1 - x^2 so that
y - 1 = +- sqrt( 1 - x^2) and
y = 1 +- sqrt(1 - x^2).
Thus for any x the limits on y are 1 - sqrt(1 - x^2) and 1 + sqrt(1 - x^2).
The expression sqrt(1 + fx^2 + fy^2) simplifies pretty nicely and the algebra doesn't get all that bad.
Note that the area of the circle x^2 + (y-1)^2 =1 is pi; the surface area of the sphere above this region will be somewhat greater than pi. The area of the entire sphere is 32 pi / 3; the area of the upper half is 16 pi / 3; the area of the upper half for y > 0 is 8 pi / 3; and it should be clear that you have less than half surface area of this part (your region doesn't contain much of the sphere near its intersection with the xy plane, where the sphere is 'steepest), so your integral should be less than 4 pi / 3. This puts the surface area of the region somewhere between pi and about 4.
19.) The portion of the surface z=9-x^2-y^2 that lies above the xy plane
I noted that when z=0 , this surface takes the form x^2+y^2=9, a circle of radius 3
My double integral will range from -3
and then from 0
Think in terms of a partition of the interval from x = -3 to 3. The 'strip' in the y direction runs from the 'bottom' of the circle to the 'top'.
You will integrate from x = -3 to x = 3, and from y = - sqrt(9-x^2) to sqrt(9 - x^2).
This surface is a paraboloid (look at xz and yz traces) because z = 9 - x^2 - y^2 has the form z = 9 - r^2, where r is distance from the z axis. The 'vertex' of the paraboloid is at (0, 0, 9).
You remember that problem I showed you about the intersection of two cylinders of radius 1 that
meet at right angles. It asks for volume, but lets say we just wanted to find the surface area of that
region. It also states: make the axis of one of the cylinders the xaxis and the other the y-axis.
well we have two equations for the cylinder, x^2+z^2=1 and then y^2 +z^2=1
What if we eliminate z from these two functions, will we just have a function in the form of x^2-
y^2=0 so would our limits on the integral vary from -1
integral with 0
x^2 - y^2 = 0 means y^2 = x^2 so that y = +- x. The cylinders intersect above and below the lines y = +- x.
The solid formed by the intersection lies above and below the square region -1 < x < 1, -1 < y < 1. The diagonals of the square correspond to the lines y = +- x. For points closer to the x than the y axis, the region is bounded above and below by the first cylinder. For points closer to the y axis than the x axis, the region is bounded above by the second cylinder.
If you wanted to find the volume of the region of intersection, you would integrate separately over the regions bounded by the two different cylinders. One region would be -1 < x < 1, -x < y < x (bounded above and below by the first cylinder). The other region would be -1 < y < 1, -y < x < y (bounded above and below by the second cylinder).
If you wanted to find the surface area you would use the same regions, integrating sqrt( 1 + fx^2 + fy^2) for the appropriate cylinder.
with f(x,y)= x^2-y^2 and then take these partials and use them in the form of sqrt(fx^2+fy^2+1)
Ill try to submit more work Saturday night."