Surface Integrals with Parameterized Surface - Part 1

The video explains how to evaluate a surface integral when the surface is given parametrically.

[ctnelsoncar]

Just wanna say thank you！This world need some more man like you

[mduduzivilakazi]

thank you so much i have a better understanding now and i love the quotes that u normally put at the end.

[MadMagzzz]

Just a tip for other viewers that in my opinion makes the last two rows of this video easier to calculate on a test where you're more stressed and likely to make some careless mistakes when substituting (from my own experience at least):

Trig identity: 2*cos(u)*sin(u) = sin(2u). Therefore:

64 * the integral of (cos(u) * sin(u)) du =
32 * the integral of (2 * cos(u) * sin(u)) du =
32 * the integral of (sin(2u)) du =
32 * (1/2 * -cos(2u)), u goes from 0 to pi/2, =
32 * 1/2 (-cos(pi) - -cos(0))=
16 (-cos(pi) + cos(0)) =
16 (cos(0) - cos(pi))
= 16 (1 - (-1)) = 16 * 2 = 32.

But in the end, it's a matter of preference of course. Thank you so much for the awesome videos on surface integrals and parameterization, something I've been struggling to get for quite some time now! =)

[arthuruppiano]

Thanks for the videos. They are well done and easy to follow.

[rovshanabdurrahimov]

thank u man. i understand everything with ur help

[dnigli]

did you HAVE to use the diagonal method when finding Ru X Rv??

[rob]

thank you, i encourage you to keep with the good work

[brandonhaag]

why is it 2cos, 2sin? why not just cos, sin?

[pranavmeelu]

isnt it supposed to be pi/4 for the first octant?

[nguyenngocbichan]

2:03 why do we know u is in the first octant?

[shefalykumari]

evaluate the surface integral of the following data.
G=cos y + sin x
S: x + y + z = 2
x>=0, y>=0, z>0