Question

Identify the surface with the given vector equation

\(r(u,v) = 2sinu i + 3cosu j + v k, \,\,0 \le v \le 2\)

Short answer

The give vector equation surface represents an elliptic cylinder

Step-by-step solution

Applied the given equation to determine x, y, and z

By using the formula,\(r(u,v) = x(u,v)i + y(u,v)j + z(u,v)k\)

\(\begin{array}{l}x = 2\sin u\\y = 3\cos u\\z = v\end{array}\)

Rewrite the x, y, and z values

Let’s rewrite the x, y, and z values

\(\begin{array}{c}\frac{x}{2} = \sin u\\\frac{y}{3} = \cos u\\z = v\end{array}\)

Solution of the vector equation

Square and add both sides

\(\begin{array}{l}{\left( {\frac{x}{2}} \right)^2} + {\left( {\frac{y}{3}} \right)^2} = {\sin ^2}u + {\cos ^2}u\\{\left( {\frac{x}{2}} \right)^2} + {\left( {\frac{y}{3}} \right)^2} = 1\end{array}\)

\({\left( {\frac{x}{2}} \right)^2} + {\left( {\frac{y}{3}} \right)^2} = 1\) represents an ellipse in xy-plane and there is no restriction on u but v is restricted and lies between 0 and 2, so, \(0 \le z \le 2\).