Question about surface Area

Question

curry · Answer

curry · Answer

pi/6 x (17^{3/2}-1) <<< was answer given

curry · Answer

@ganeshie8 @dan815

curry · Answer

i just did ||Tu x Tv|| du dv. but that gave me u(sqrt(5)) insided the final integral.

curry · Answer

so it doesn't lead me to the correct answer, but isn't it just supposed to be that?

curry · Answer

wait why does yours have sqrt 1 + u^2?

curry · Answer

and so i guess my main question is, regardless of whether i'm finding an area over a graph g(x,y), or i'm just finding the area, i can use just ||Tu x Tv||?

ganeshie8 · Answer

Oh sorry i used the surface from previous problem itself :  (ucosv, usinv, u)
which is wrong

jhannybean · Answer

Oh don't you have to use Stokes Theorem to find area of the surface??

ganeshie8 · Answer

you're getting this right http://www.wolframalpha.com/input/?i=%5Cint_0%5E%7B2pi%7D%5Cint_0%5E2+%28sqrt%285u%5E4%29%29+dudv

curry · Answer

wait wait,

curry · Answer

so sqrt(4u^4 + u^2) right?

curry · Answer

and my main question is still there, regardless of whether i'm finding an area over a graph g(x,y), or i'm just finding the area, i can use just ||Tu x Tv||?

jhannybean · Answer

$$\int\int_s \vec F \cdot dS = \int\int_D \vec F \cdot (\vec r_u \times \vec r_v)dA$$

curry · Answer

but you're not always given vector field...

curry · Answer

hence that formula doesn't always apply.

jhannybean · Answer

$$\vec r(u,v) = \langle u\cos(v), u\sin(v), u^2\rangle$$ Can you not use this to find the vector field?

curry · Answer

there is no vecctor field, it's just finding surface area, hahaha

curry · Answer

@ganeshie8, i think our sqrts aren't matching up, i got qrt(4u^4 + u^2)

ganeshie8 · Answer

oh yours is right, forget about my link..

curry · Answer

oo, kk, and so my question about the surface and graphs?

ganeshie8 · Answer

g(x,y) = 1 when you're just finding the area of surface

jhannybean · Answer

hmm... I thought we could use our parameters to find the function in the form z=f(x,y)!! :(

ganeshie8 · Answer

think of finding the area of given surface  as : ` integrating the function g(x,y) = 1 over the given surface`

ganeshie8 · Answer

Recall how $\text{area } = \iint~ 1~ dA $

this gives area of any shape in xy plane

jhannybean · Answer

$$\left|\vec r_u \times \vec r_v\right| = \left|\langle \cos(v),\sin(v), 2u\rangle \times \langle -usin(v), u\cos(v),0\rangle\right| $$

jhannybean · Answer

no?

ganeshie8 · Answer

i got the same too jhanny

dan815 · Answer

they keep giving u freaky surfaces

ganeshie8 · Answer

this looks like a paraboloid

dan815 · Answer

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