Question

find the area of the part of paraboloid x=4y^2+4z^2 that lies inside the cylinder z^2+y^2=3
@Hero @ganeshie8

Answer 1

where are u stuck ?

Answer 2

just the axes were changed... other than that the problem is bit straightforward right ?

Answer 3

Surface area = \[ \iint \limits_R \sqrt{f_y ^2 + f_z^2 + 1} ~dy dz\]

Answer 4

take the partials and plugin

Answer 5

Surface area = \[ \iint \limits_R \sqrt{(8y)^2 + (8z)^2 + 1} ~dy dz\]

\[ \iint \limits_R \sqrt{64(y^2+z^2) + 1} ~dy dz\]

Answer 6

next look at the cylinder to setup the bounds

Answer 7

z^2 + y^2 <= 3

gives a shodow of disk of radius sqrt(3) in yz plane

Answer 8

Surface area = \[ \iint \limits_R \sqrt{(8y)^2 + (8z)^2 + 1} ~dy dz\]

\[ \iint \limits_R \sqrt{64(y^2+z^2) + 1} ~dy dz\]

\[ \int \limits_0^{2\pi} \int \limits_0^{\sqrt{3}} \sqrt{64(r^2) + 1} ~r ~dr d\theta\]

Answer 9

evaluate

Answer 10

how did you plug it in to wolffram aplha last time?

Answer 11

http://www.wolframalpha.com/input/?i=%5Cint++%5Climits_0%5E%7B2%5Cpi%7D++%5Cint+%5Climits_0%5E%7B%5Csqrt%7B3%7D%7D++%5Csqrt%7B64%28r%5E2%29+%2B+1%7D+r+dr+d%5Ctheta

Answer 12

thannks