Because I've gone ahead with trying to parameterize directly and learned the hard way that the resulting integral is large and annoying to work with, I'll propose a less direct approach.
Rather than compute the surface integral over straight away, let's close off the hemisphere with the disk of radius 9 centered at the origin and coincident with the plane . Then by the divergence theorem, since the region is closed, we have
where is the interior of . has divergence
so the flux over the closed region is
The total flux over the closed surface is equal to the flux over its component surfaces, so we have
Parameterize by
with and . Take the normal vector to to be
Then the flux of across is
Answer:
120strip
Step-by-step explanation:
Answer:
HERE IS YOUR ANSWER
Step-by-step explanation:
Use the mirror equation:
1/di + 1/do = 1/f
where di = -10 cm and f = +15 cm. (Note that di is negative if the image is virtual.)
Substitute and solve for do.
1/do + 1/(-10 cm) = 1/(15 cm)
1/do = 1/(15 cm) - 1/(-10 cm) = 5/(30 cm)
do = 6 cm
Hope it helps you
Regards,
Rachana
2 and 12 both correspond with angle 5
Its is something give me more detail