00:01
Hi, from the question given that here we need to evaluate the surface integral, double integral over the region s, x, y, z, ds, where s is the cone with parametric equation x is equal to u cos v and y is equal to u sin v and z is equal to u, where 0 is less than or equal to u is less than or equal to 1 and 0 is less than or equal to v is less than or equal to pi over 2.
00:35
Now we have the parametric equation r of u, v is equal to inner product of u cos v u sin v u.
00:47
Now r u that is partial derivative will be inner product of cos v sin v 1.
01:00
Now r v is equal to inner product of minus u sin v u cos v 0.
01:12
Now r u cross r v is equal to i j k cos v sin v 1 minus u sin v u cos v 0.
01:33
So magnitude of r u cross r v is equal to under root of u squared cos squared v plus sin squared v plus u squared.
01:54
So this is equal to u root 2.
01:57
Now the required surface area will be triple double integral over the region s, x, y, z, ds.
02:05
So this could be equal to integral 0 to 1 that is u varies from 0 to 1 and v varies from 0 to pi by 2 u cube cos v sin v u root 2 dv du.
02:28
So this is equal to root 2 integral 0 to 1 integral 0 to pi by 2 u power 4 cos v sin v dv du...