Area of part of parametric function I need to get area of function: $x= 2\sqrt{2}\cos ^3 t$ and $y= 4\sqrt{2}{\sin ^3 t}$, but only the part when $x\geq1$. How can I do that? I know that area of full function would be
$$S= \int_a^b y(t)x'(t) \, dt $$
$a \leq t \leq b$
 A: UPDATE. The area from $x=1$ to $x(0)=a=2\sqrt{2}$ enclosed by the parametric curve 
\begin{equation*}
\left( x(t),y(t)\right) =\left( 2\sqrt{2}\cos ^{3}t,4\sqrt{2}\sin
^{3}t\right) 
\end{equation*}
is 
\begin{equation*}
S=2\int_{1}^{a}y\,dx.
\end{equation*}

Since $x^{\prime }(t)=-6\sqrt{2}\cos ^{2}t\sin t$ and
\begin{equation*}
x(t)=1\Rightarrow 2\sqrt{2}\cos ^{3}t=1\Leftrightarrow t_{1}=\frac{\pi }{4}
,t_{2}=-t_{1},
\end{equation*}
the same area, using your formula, can be expressed as
\begin{eqnarray*}
S &=&2\int_{\pi /4}^{0}\left( 4\sqrt{2}\sin ^{3}t\right) \left( -6\sqrt{2}%
\cos ^{2}t\sin t\right) \,dt \\
&=&96\int_{0}^{\pi /4}\cos ^{2}t\sin ^{4}t\,dt \\
&=&96\int_{0}^{\pi /4}\left( 1-\sin ^{2}t\right) \sin ^{4}t\,dt \\
&=&96\int_{0}^{\pi /4}\sin ^{4}t\,dt-96\int_{0}^{\pi /4}\sin ^{6}t\,dt \\
&=&\cdots  \\
&=&-2+\frac{3}{2}\pi  \\
&\approx &2.7124.
\end{eqnarray*}

Warning. This answer is no longer updated. The question was changed to the parametric curve $$(x= 2\sqrt{2}\cos ^3 t,y= 4\sqrt{2}{\sin ^3 t})$$

The area from $x=1$ to $x(0)=a=2\sqrt{2}$ enclosed by the parametric curve 
$$\left( x(t),y(t)\right) =\left( 2\sqrt{2}\cos t^{3},4\sqrt{2}\sin
t^{3}\right) $$
is given by
\begin{equation*}
S=2\int_{1}^{a}y\,dx.
\end{equation*}

Since $x^{\prime }(t)=-6\sqrt{2}t^{2}\sin t^{3}$ and
\begin{equation*}
x(t)=1\Rightarrow 2\sqrt{2}\cos t^{3}=1\Leftrightarrow t_{1}=\arccos ^{1/3}
\frac{\sqrt{2}}{4},t_{2}=-t_{1},
\end{equation*}
the same area, using your formula, can be expressed as
\begin{eqnarray*}
S &=&2\int_{t_{1}}^{0}\left( 4\sqrt{2}\sin t^{3}\right) \left( -6\sqrt{2}
t^{2}\sin t^{3}\right) \,dt \\
&=&96\int_{0}^{t_{1}}t^{2}\sin ^{2}t^{3}\,dt=96\left[ \frac{1}{6}\left(
t^{3}-\cos t^{3}\sin t^{3}\right) \right] _{0}^{t_{1}} \\
&=&16\left. \left( t^{3}-\cos t^{3}\sin t^{3}\right) \right\vert
_{0}^{\arccos ^{1/3}\frac{\sqrt{2}}{4}} \\
&=&16\left( \arccos \frac{\sqrt{2}}{4}-\frac{\sqrt{2}}{4}\sin \left( \arccos 
\frac{\sqrt{2}}{4}\right) \right)  \\
&=&-2\sqrt{7}+16\arccos \frac{\sqrt{2}}{4} \\
&\approx &14.059.
\end{eqnarray*}
