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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

$$S= \int_a^b y(t)x'(t) \, dt $$ $a \leq t \leq b$

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"Only the part when $x\geq 1$." Do you mean $t \geq 1$? –  Vladhagen Apr 7 at 16:51
    
No that's the confusing part. Text: Find area bounded by param. funct., x=1, x>=1. No further explanations. –  Kristians Kuhta Apr 7 at 16:58
    
You will need to find the intervals of $t$ in which $\cos t^3$ is greater than $\frac{1}{\sqrt{2}}$. Then integrate over those intervals. –  Vladhagen Apr 7 at 18:51
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Isn't it $\frac{1}{2\sqrt{2}}$? Isn't this right? $1\leq x \leq 2\sqrt{2}$, $1\leq2\sqrt{2}\cos{t}^3\leq 2\sqrt{2}$ –  Kristians Kuhta Apr 7 at 19:06

1 Answer 1

up vote 2 down vote accepted

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*}

enter image description here

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 This answer is no longer updated.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*}

enter image description here

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*}

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Sorry I don't know yet how to use LaTex right those cos[t] functions are cubed, not the argument t itself. –  Kristians Kuhta Apr 7 at 19:30
    
@KristiansKuhta Do you mean $x= 2\sqrt{2}\cos^3(t); y= 4\sqrt{2}\sin^3(t)$? (`x= 2\sqrt{2}\cos^3(t); y= 4\sqrt{2}\sin^3(t)) –  Américo Tavares Apr 7 at 19:33
    
Yes, and again sorry about that. –  Kristians Kuhta Apr 7 at 19:36
    
@KristiansKuhta Try to evaluate the new integral using a similar approach. –  Américo Tavares Apr 7 at 19:41
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@KristiansKuhta I've updated the answer. –  Américo Tavares Apr 7 at 20:32

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