Question

Find the area of the surface.

The part of the plane with vector equation \(r\left( {u,v} \right) = \left\langle {u + v,2 - 3u,1 + u - v} \right\rangle \) that is given by \(0 \le u \le 2, - 1 \le v \le 1\) .

Short answer

The required surface area is \(4\sqrt {22} \) .

Step-by-step solution

Given

Consider the vector equation of the plane.

\(r\left( {u,v} \right) = \left\langle {u + v,2 - 3u,1 + u - v} \right\rangle \)where \(0 \le u \le 2, - 1 \le v \le 1\)

Then

\(\begin{array}{l}{r_u}\left( {u,v} \right) = \left\langle {1, - 3,1} \right\rangle \\{r_v}\left( {u,v} \right) = \left\langle {1,0, - 1} \right\rangle \end{array}\)

\(\begin{array}{c}\left\langle {1, - 3,1} \right\rangle \times \left\langle {1,0, - 1} \right\rangle = \left\langle { - 3\left( { - 1} \right) - 0\left( 1 \right), - 1\left( 1 \right) - 1\left( 1 \right),1\left( 0 \right) - 1\left( { - 3} \right)} \right\rangle \\ = \left\langle {3, - 2,3} \right\rangle \end{array}\)

Next, find the length of this cross-product vector.

\(\left| {{r_u} \times {r_v}} \right| = \sqrt {{3^2} + {{\left( { - 2} \right)}^2} + {3^2}} = \sqrt {22} \)

Finally, integrate the length over the domain region, \(D = \left\{ {\left( {u,v} \right)|0 \le u \le 2, - 1 \le v \le 1} \right\}\) .

Calculating surface area

Work with the inner integral first.

\(\begin{array}{c}\int\limits_0^2 {\sqrt {22} du} = \sqrt {22} \left[ u \right]_0^2\\ = \sqrt {22} \left( {2 - 0} \right)\\ = 2\sqrt {22} \end{array}\)

Next, substitute this in and evaluate the outer integral.

\(\begin{array}{c}\int\limits_{ - 1}^1 {2\sqrt {22} du} = 2\sqrt {22} \left[ v \right]_{ - 1}^1\\ = 2\sqrt {22} \left( {1 - \left( { - 1} \right)} \right)\\ = 4\sqrt {22} \end{array}\)

Therefore, the surface area of the plane below is \(4\sqrt {22} \) .

Graph

The diagram of the plane \(r\left( {u,v} \right) = \left\langle {u + v,2 - 3u,1 + u - v} \right\rangle \) where \(0 \le u \le 2, - 1 \le v \le 1\) is shown below: