Question

Find the area of the surface. The surface with parametric equations x = u2, y = uv, z = 1 2 v2, 0 ≤ u ≤ 2, 0 ≤ v ≤ 2.

Answer 1

`\int\limits^2_0 {} \, \int\limits^2_0 \sqrt{576v^(4)+2304u^(2)v^(2)+4u^(4)}`du dv

Explanation:

let, r = x i + y j + z k , where i, j, k are unit vectors.

r =
`u^(2)` i + uv j + 12
`v^(2)` k

we know that the surface area of a surface represented by r(u,v) is

=
`\int\limits^2_0 {} \, \int\limits^2_0 {(dr)/(du) (cross)(dr)/(dv) } \, dudv`

here,

`(dr)/(du)` = 2u i + v j

`(dr)/(dv)` = u j + 24 v k

Cross product =
`\left[\begin{array}{ccc}i&j&k\\2u&v&0\\0&u&24v\end{array}\right]`

= 24
`v^(2)` i - 48 uv j + 2
`u^(2)` k

The modulus of the cross product is
`\sqrt{576v^(4)+2304u^(2)v^(2)+4u^(4) }`

so, the surface area is

`\int\limits^2_0 {} \, \int\limits^2_0 \sqrt{576v^(4)+2304u^(2)v^(2)+4u^(4)}`du dv

and the answer has to be left as the integral itself as the integral of square root of biquadratic can not be calculated(with random co efficients).