Consider a surface that is parametrised by r(x,y)=(xyx2y) \mathbf{r}(x,y) = \left(\begin{array}{c} x \ y \ x^2 y \end{array} ight) .The surface area element is equal to

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Consider a surface that is parametrised by r(x,y)=(xyx2y) \mathbf{r}(x,y) = \left(\begin{array}{c} x \ y \ x^2 y \end{array}ight) .The surface area element is equal to

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We begin by identifying the parametrization r(x,y) = (x, y, x^2 y). To obtain the surface area element, we compute the tangent vectors partial derivatives: r_x = ∂r/∂x = (1, 0, 2xy) and r_y = ∂r/∂y = (0, 1, x^2).

Next, we take their cross product r_x × r_y, which gives a vector normal to the surface with magnitude equal to the differential area element: r_x × r_y = |i   j   k; 1   0  2xy; 0   1  x^2| = (-2xy, -x^2, 1).

The area element is the norm of this cross product times dx dy: dS = |r_x × r_y| dx dy. Compute the norm: |(-2xy, -x^2, 1)| = sqrt{(-2xy)^2 + (-x^2)^2 + 1^2} = sqrt{4x^2y^2 + x^4 + 1}.

Therefore, the surface area element is dS = sqrt{4x^2 y^2 + x^4 + 1} \, dx \, dy. This matches the given expression exactly.

Consider a surface that is parametrised by r(x,y)=(xyx2y) \mathbf{r}(x,y) = \left(\begin{array}{c} x \\ y \\ x^2 y \end{array}\right) .The surface area element is equal to单项选择题

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