I need to find the area between two curves:

$$\begin{cases} x=\sqrt { 2 } \cos { t } \\ y=4\sqrt { 2 } \sin { t } \end{cases}\\ y=4\quad (y\ge 4)$$

I came up with:

$$\frac { 1 }{ 4 } \left( 8\pi n +\pi \right) \le t\le \frac { 1 }{ 4 } \left( 8\pi n + 3\pi \right) $$

$$\int _{ \frac { 1 }{ 4 } (8\pi n+\pi ) }^{ \frac { 1 }{ 4 } (8\pi n+3\pi ) }{ 4\sqrt { 2 } \sin { (t) } } -\sqrt { 2 } \cos { (t) } \quad dx =2\left( \sin { (2\pi n) } +4\cos { (2\pi n) } \right)$$

So the area is a function of $n$, though I was supposed to get a finite solution. What am I doing wrong?

  • $\begingroup$ Restrict your angle to have one full revolution, and then you can find corresponding angles that might fit your restriction on $y$. Right now you're not restricting your angles at all and draw the ellipse infinitely many times. $\endgroup$ – Kaster Apr 14 '16 at 20:05
  • $\begingroup$ @Kaster I don't quite understand why that is legal. Can you elaborate a bit? I haven't done any calculus for 4 years. $\endgroup$ – Eli Korvigo Apr 14 '16 at 20:51
  • $\begingroup$ If you plot your curve according the the parametric equation you're given and do not restrict your angle $t$, you will draw same ellipse over and over again, and your final curve will contain same point infinitely many times, which is redundant. So it's not the matter of legality, but simply having a function so you can use integration. $\endgroup$ – Kaster Apr 14 '16 at 20:58
  • $\begingroup$ How is your answer a function of $n$ ? What you wrote is $8$. (Anyway, your integral is wrong.) $\endgroup$ – Yves Daoust Apr 14 '16 at 21:25
  • $\begingroup$ @YvesDaoust you are right, the answer is not a function of $n$, because $\sin 2\pi n=0$ for any $n$. I knew there were errors in the solution, that was the sole reason to post the question. Now I know what to revise. $\endgroup$ – Eli Korvigo Apr 14 '16 at 22:02

As I said in my comment, you need to restrict your angles, both in general, to have only one revolution around the origin, and when you want to meet your additional restriction on $y$. For the first case, it's obvious that $t \in [0, 2\pi)$. For the second one, solve an easy equation $$ y = 4 \implies 4\sqrt 2 \sin t = 4 \implies \sin t = \frac 1{\sqrt 2} \implies t_1 = \frac \pi 4,\ t_2 = \frac {3\pi}4 $$

If you visualize the analysis above, you get


You need to find the area of the shape with red dome and black straight base,

Finding the area of the red curve and $y = 0$ is as easy as $$ A_f = \int_{\frac {3\pi} 4}^{\frac \pi 4} y(t)\ x'(t)\ dt = 8 \int_{\frac \pi 4}^{\frac {3\pi}4} \sin^2 t\ dt = 4 \left . \left( t - \frac {\cos 2t}2\right) \right |_{\frac \pi 4}^{\frac {3\pi}4} = 2(2 + \pi) $$ since you know your angles.

And the area you need is the difference between the area above and rectangle with dashed sides, solid black top and bottom side on $x$ axis, which is $$ A_r = 2 \cdot 4 = 8 $$ and finally, $A = A_f - A_r = 2\pi - 4$


In integration, I used the positive direction of $x$ to put upper and lower bounds for $t$, but then because of the negative sign that comes from $x'(t)$ I switched them again to get increasing order of angles.

  • $\begingroup$ Thank you for the elaborate explanation. $\endgroup$ – Eli Korvigo Apr 14 '16 at 22:03
  • 1
    $\begingroup$ Why are you getting a triangle for $A_t$, not a rectangle? Notice that if you blew the figure up by a factor of $4$ in the $x$-direction, you wouldn't need any calculus. You would have a sector of area $\frac12(4\sqrt2)^2\frac{\pi}2=8\pi$ less a triangle of area $\frac12(8)(4)=16$, for a total $8\pi-16$, then divide by $4$ because the figure was $4\times$ bigger in the $x$-direction, so you end up with $A=2\pi-4$. $\endgroup$ – user5713492 Apr 15 '16 at 1:26
  • $\begingroup$ @user5713492 you're absolutely right. My mistake. I was thinking about something else. Thanks. $\endgroup$ – Kaster Apr 15 '16 at 1:41
  • $\begingroup$ @EliKorvigo please see corrected answer. Sorry. $\endgroup$ – Kaster Apr 15 '16 at 1:41

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