What is surface area of intersection of two cylinders: x²+z² < 1 and y²+z²<1?

Answer 1

Hello

The intersection of these 2 cylinders of radii 1 is called a Steinmetz solid and the surface area is equal to 16r^2 (calculated by Archimedes himself) Therefore the surface area is equal to 16.

You have x^2 + z^2 = 1 and y^z + z^2 = 1, you get
x = +- y and
z= sqrt(1-y^2)

Surface area is calculated by integrating a segment of a curve length and the width.
Which equals Integral x ds.
ds = sqrt ( 1 + (dz/dy)^2 ) dy
= 1/sqrt(1-y^2) dy

integral of x 1/sqrt(1-y^2) dy = integral of y/sqrt(1-y^2) dy
Setting the ranges to 0 to 1 will give you the 1/16 of the surface area due to symmetry.
-sqrt(1-y^2) evaluated at 0 and 1.
0 + (1) = 1
1*16 = 16.

Answer 2

We have to integrate dx*dy*dz over the enclosed volume.

-1 < z < 1
-sqrt(1-z^2) < x < +sqrt(1-z^2)
-sqrt(1-z^2) < y < +sqrt(1-z^2)

After performing the x and y integrations, you end up having to integrate
4*(1-z^2) dz between -1 and +1
to get 16/3 cu units

Oh damn, you wanted surface area.

This was a really fun piece of work.
Let's move from y = 0 to 1. For each interval dy, we wish to find the arclength of intersection.
On the cylinder x^2 + z^2 = 1, the intersection is clearly a circular arc, but the enclosed angle varies with y.

x^2 + z^2 = 1
intersects
y^2 + z^2 = 1
at x = +/- y

So on the arc x^2 + z^2 = 1, the limits of the intersection are
x = y, z = sqrt(1-y^2)
and x = -y, z = sqrt(1-y^2)
The angle enclosed = 2*arccos(y)
The arclength = 2*r*arccos(y) = 2*arccos(y)
The elemental area, dA = 2*y*arccos(y)

Now we must integrate this from y = 0 to 1
A = 2*[y*arccos(y) - sqrt(1-y^2)] | 0 to 1
= 2

And there are 8 such areas.
Total surface area of intersection = 16 sq units.

Answer 3

well, surface area is double integral sqrt((z'x)^2+(z'y)^2+1) dA here when z=1, x^2+y^2=1 and z=2, x^2+y^2=2, so we can use polar coordinates. r goes from the smaller radius which is 1 to the bigger which is sqrt(2). Tetha goes from 0 to 2pi. next z'x= 2x, (z'x)^2= 4x^2, z'y=2y, (z'y)^2=4y^2. so we get double integral sqrt(1+4x^2+4y^2) r dr dtetha now replace the 4x^2+4y^2 by 4r^2. we get double integral sqrt(1+4r^2)r dr dtetha use u substitution, u=1+4r^2,du=8r dr, du/8= rdr so double integral (1/8) u^1/2 du dtetha, = 1/8 integral 2/3 (1+4r^2)^3/2 from 1 to sqrt(2) dthetha = 2/24 integral (27-5^3/2) dtetha = 1/12 *(27-5^3/2) thetha from 0 to 2pi= 1/12 * 2pi* (27-5^3/2)= pi/6(27-5^3/2)

Answer 4

these are actually circle, you need 3 variables describe a cylinder. well if u meant circle then the intersection is a line tht is greater then -1 but less then 1.

Answer 5

Actually, what you have described are two circles, one in the x-z plane and one in the y-z plane.

Picture two circles at right angles to one another. Their intersection would form a line segment. In fact, in this case it is the open interval (-1,1) on the z-axis.