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

Answer:

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

Step-by-step 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

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

here,

      $\frac{dr}{du}$ = 2u i + v j

      $\frac{dr}{dv}$ = 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).