Find the area of the part of the plane with vector equation
r(u, v) =
<2 + v, u − 7v, 5 − 4u + v>
that is given by
0 ≤ u ≤ 1, 0 ≤ v ≤ 1.
To find the area of the part of the plane with the given vector equation, we need to find the area of the corresponding parallelogram.
In general, the area of a parallelogram with sides defined by vectors A and B is given by the magnitude of the cross product of A and B:
Area = |A × B|
In this case, we can find the sides of the parallelogram by taking two partial derivatives of the vector equation with respect to u and v, and then taking the cross product of the resulting vectors.
Taking the partial derivative of r(u,v) with respect to u, we get:
∂r/∂u = <1, -4, -4>
Taking the partial derivative of r(u,v) with respect to v, we get:
∂r/∂v = <1, 1, 1>
Now, we can take the cross product of ∂r/∂u and ∂r/∂v to find the area:
∂r/∂u × ∂r/∂v = <1, -4, -4> × <1, 1, 1>
= <8, -8, -8>
Taking the magnitude of this cross product vector:
|∂r/∂u × ∂r/∂v| = √(8^2 + (-8)^2 + (-8)^2)
= √(64 + 64 + 64)
= √192
= 8√3
So, the area of the part of the plane given by the vector equation is 8√3 square units.
To find the area of the given part of the plane, we will use the concept of surface area. The surface area of a parametrically defined surface in a particular region can be found using a double integral.
In this case, the vector equation r(u, v) = <2 + v, u − 7v, 5 − 4u + v> defines the surface in three-dimensional space. The bounds for u and v are 0 ≤ u ≤ 1 and 0 ≤ v ≤ 1, respectively.
To calculate the area, we need to set up a double integral using the magnitude of the cross product of the partial derivatives of r(u, v) with respect to u and v. The magnitude of the cross product gives us the local area element associated with an infinitesimally small change in both u and v.
Let's start by finding the partial derivatives of r(u, v) with respect to both u and v:
∂r/∂u = <1, -7, -4>
∂r/∂v = <1, 1, 1>
Next, we will calculate the cross product of these two partial derivatives:
∂r/∂u x ∂r/∂v = (1i - 7j - 4k) x (1i + 1j + 1k)
= (11i + 5j + 8k)
Now we have the magnitude of the cross product:
|∂r/∂u x ∂r/∂v| = √(11^2 + 5^2 + 8^2)
= √210
Finally, we set up the double integral to find the area:
Area = ∫∫ |∂r/∂u x ∂r/∂v| dA
= ∫[0,1]∫[0,1] √210 du dv
Evaluating this double integral will give us the desired area.