If a parametric surface given by ri(u, v) f(,v)i+ g(u, v h (u, v)k and 1 S u S -2 S v S 2 has surface area equal to 2, what is the surface area of the parametric surface given by rz (u,v) 2r1(u, v) with -1 S u S 1-2 S v S 2?
The Correct Answer and Explanation is :
The problem provides two parametric surfaces, ( \mathbf{r}_1(u, v) ) and ( \mathbf{r}_2(u, v) = 2\mathbf{r}_1(u, v) ), and asks for the surface area of the surface described by ( \mathbf{r}_2(u, v) ) when the surface area of ( \mathbf{r}_1(u, v) ) is given as 2.
Solution:
Given that the surface area of the parametric surface ( \mathbf{r}_1(u, v) ) is ( A_1 = 2 ), we know the formula for the surface area of a parametric surface:
[
A = \iint_{D} \left| \frac{\partial \mathbf{r}}{\partial u} \times \frac{\partial \mathbf{r}}{\partial v} \right| \, du \, dv
]
Now consider the parametric surface ( \mathbf{r}_2(u, v) = 2\mathbf{r}_1(u, v) ). We want to find the surface area of ( \mathbf{r}_2(u, v) ). The key observation is that scaling the surface by a factor of 2 will affect the area. To see why, note that:
- ( \mathbf{r}_2(u, v) = 2\mathbf{r}_1(u, v) ), so the derivatives of ( \mathbf{r}_2 ) with respect to ( u ) and ( v ) are:
[
\frac{\partial \mathbf{r}_2}{\partial u} = 2 \frac{\partial \mathbf{r}_1}{\partial u}, \quad \frac{\partial \mathbf{r}_2}{\partial v} = 2 \frac{\partial \mathbf{r}_1}{\partial v}
]
- The cross product ( \frac{\partial \mathbf{r}_2}{\partial u} \times \frac{\partial \mathbf{r}_2}{\partial v} ) will therefore be:
[
\frac{\partial \mathbf{r}_2}{\partial u} \times \frac{\partial \mathbf{r}_2}{\partial v} = 4 \left( \frac{\partial \mathbf{r}_1}{\partial u} \times \frac{\partial \mathbf{r}_1}{\partial v} \right)
]
- The magnitude of the cross product scales by a factor of 4 (because ( 2 \times 2 = 4 )).
Thus, the surface area of ( \mathbf{r}_2 ) will be:
[
A_2 = 4 \times A_1 = 4 \times 2 = 8
]
Conclusion:
The surface area of the parametric surface given by ( \mathbf{r}_2(u, v) = 2\mathbf{r}_1(u, v) ) is ( \boxed{8} ).
Explanation:
The surface area formula involves the magnitude of the cross product of the partial derivatives of the position vector with respect to ( u ) and ( v ). When the position vector is scaled by a factor of 2, both partial derivatives are scaled by 2. The magnitude of the cross product, which determines the infinitesimal area element, scales by the square of this factor (in this case, ( 2^2 = 4 )). Hence, the surface area increases by a factor of 4, resulting in a surface area of 8.