\( \DeclareMathOperator{\abs}{abs} \newcommand{\ensuremath}[1]{\mbox{$#1$}} \)

Find the area inside the loop given by r(t)=<sin(Pi*t),t-t^4> for 0 ≤ t≤ 1, as shown below.

(%i4) wxplot2d( [parametric, sin(t·%pi),t−t^4], [t,0,1], [color,black]);
\[\tag{%t4} \] (Graphics)
\[\tag{%o4} \]

First we use Green's Theorem. Let F = <M,N> be the vector field given by N = x and M=0. Now area = int int (1) dA = int int (dN/dx - dM/dy) dA = (by Green's Theorem) int M dx + N dy =  int x(t) y'(t) dt for t from 0 to 1.

Since y(t) = t - t^4, we get that y'(t) = 1 - 4t^3.

(%i6) integrate(sin(%pi·t)·(1−4·t^3),t,0,1);
\[\tag{%o6} \frac{1}{\ensuremath{\pi} }-\frac{3 {{\ensuremath{\pi} }^{2}}-24}{{{\ensuremath{\pi} }^{3}}}\]
(%i7) float(%);
\[\tag{%o7} 0.1374170540292064\]

Using Calc I methods you would need to define y1(x) for the top curve and y2(x) for the bottom curve and then integrate y1(x)-y2(x) for x = 0 to 1.

This is done below. See if you can figure out how to find y1 and y2. (Finding y2 is easier.)

(%i12) y1(x) := 1−1/%pi·asin(x)−(1−1/%pi·asin(x))^4;
\[\tag{%o12} \operatorname{y1}(x):=1-\frac{1}{\ensuremath{\pi} } \operatorname{asin}(x)-{{\left( 1-\frac{1}{\ensuremath{\pi} } \operatorname{asin}(x)\right) }^{4}}\]
(%i13) y2(x) := 1/%pi·asin(x)−(1/%pi·asin(x))^4;
\[\tag{%o13} \operatorname{y2}(x):=\frac{1}{\ensuremath{\pi} } \operatorname{asin}(x)-{{\left( \frac{1}{\ensuremath{\pi} } \operatorname{asin}(x)\right) }^{4}}\]
(%i14) wxplot2d([y1(x),y2(x)],[x,0,1],[color,red,blue]);
\[\tag{%t14} \] (Graphics)
\[\tag{%o14} \]
(%i17) integrate(y1(x)−y2(x),x,0,1);
\[\tag{%o17} -\frac{2 {{\ensuremath{\pi} }^{2}}-24}{{{\ensuremath{\pi} }^{3}}}\]
(%i18) float(%);
\[\tag{%o18} 0.1374170540292064\]
Created with wxMaxima.