This is way off. To compute area you need only a single integral or at most a double integral.kougou said:Homework Statement
Use spherical cords to compute area of a disk, that's center at x,y=0, and z=4, having a radius of 3
Homework Equations
I set up a triple ∫∫∫ (r^2* sin(theta)), running from phi =0 to 2pi,
theta=0 to arcsin(3/5), r=5sin(theta) to 5.
kougou said:The Attempt at a Solution
It doesn't give me a current answer. What's wrong
Am I even on the right track?
The use of spherical cord allows for a more accurate and precise calculation of the area of a disk. By wrapping the cord around the disk, it takes into account the curvature of the disk and provides a more comprehensive measurement.
Yes, the spherical cord method can be used for any size or shape of a disk as long as the cord is able to wrap around it completely. It is not limited to just circular disks, but can also be used for irregularly shaped disks.
The length of the spherical cord can be measured by simply wrapping it around the disk and marking the starting and ending points. This length can then be measured using a ruler or measuring tape.
No, the spherical cord method is specifically designed for calculating the area of a 2-dimensional disk. It may not provide accurate results for 3-dimensional objects.
One limitation of the spherical cord method is that it may not provide precise measurements for disks with very small or large radii. Additionally, the method may not be feasible for objects with irregular surfaces or those that cannot be easily wrapped with a cord.