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.
Accepted Solution
A:
Answer: [tex]\int\limits^2_0 {} \, \int\limits^2_0 \sqrt{576v^{4}+2304u^{2}v^{2}+4u^{4}}[/tex]du dvStep-by-step explanation:let, r = x i + y j + z k , where i, j, k are unit vectors.r = [tex]u^{2}[/tex] i + uv j + 12 [tex]v^{2}[/tex] kwe know that the surface area of a surface represented by r(u,v) is = [tex]\int\limits^2_0 {} \, \int\limits^2_0 {\frac{dr}{du} (cross)\frac{dr}{dv} } \, dudv[/tex]here, [tex]\frac{dr}{du}[/tex] = 2u i + v j [tex]\frac{dr}{dv}[/tex] = u j + 24 v k Cross product = [tex]\left[\begin{array}{ccc}i&j&k\\2u&v&0\\0&u&24v\end{array}\right][/tex] = 24 [tex]v^{2}[/tex] i - 48 uv j + 2 [tex]u^{2}[/tex] kThe modulus of the cross product is [tex]\sqrt{576v^{4}+2304u^{2}v^{2}+4u^{4} }[/tex]so, the surface area is [tex]\int\limits^2_0 {} \, \int\limits^2_0 \sqrt{576v^{4}+2304u^{2}v^{2}+4u^{4}}[/tex]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).