Here is a parameterization for this curve.

\[\vec r\left( t \right) = \left\langle {3\cos \left( t \right), - 3\sin \left( t \right)} \right\rangle \,\,\,\,\,\,\,\,0 \le t \le \pi \]

We could also break this up into parameter form as follows.

\[\begin{array}{*{20}{l}}{x = 3\cos \left( t \right)}\\{y = - 3\sin \left( t \right)}\end{array}\hspace{0.25in}0 \le t \le \pi \]

Either form of the parameterization will work for the problem but we’ll use the vector form for the rest of this problem.

Show Step 3

We’ll need the magnitude of the derivative of the parameterization so let’s get that.

\[\begin{align*}\vec r'\left( t \right) & = \left\langle { - 3\sin \left( t \right), - 3\cos \left( t \right)} \right\rangle \\ \left\| {\vec r'\left( t \right)} \right\| &= \sqrt {{{\left( { - 3\sin \left( t \right)} \right)}^2} + {{\left( { - 3\cos \left( t \right)} \right)}^2}} \\ & = \sqrt {9{{\sin }^2}\left( t \right) + 9{{\cos }^2}\left( t \right)} = \sqrt {9\left( {{{\sin }^2}\left( t \right) + {{\cos }^2}\left( t \right)} \right)} = \sqrt 9 = 3\end{align*}\]

We’ll also need the integrand “evaluated” at the parameterization. Recall all this means is we replace the \(x\)/\(y\) in the integrand with the \(x\)/\(y\) from parameterization. Here is the integrand evaluated at the parameterization.

\[2y{x^2} - 4x = 2\left( { - 3\sin \left( t \right)} \right){\left( {3\cos \left( t \right)} \right)^2} - 4\left( {3\cos \left( t \right)} \right) = - 54\sin \left( t \right){\cos ^2}\left( t \right) - 12\cos \left( t \right)\] Show Step 4

The line integral is then,

\[\begin{align*}\int\limits_{C}{{2y{x^2} - 4x\,ds}} & = \int_{0}^{\pi }{{\left( { - 54\sin \left( t \right){{\cos }^2}\left( t \right) - 12\cos \left( t \right)} \right)3\,dt}}\\ & = 3\left. {\left[ {18{{\cos }^3}\left( t \right) - 12\sin \left( t \right)} \right]} \right|_0^\pi = \require{bbox} \bbox[2pt,border:1px solid black]{{ - 108}}\end{align*}\]