0
$\begingroup$

I need to find the line integral of $x dy$ where the curve is $x^2 +y^2 =a^2$ I know that if I take $x=(a^2-y^2)^\frac(1/2)$ for the semicircle right of $x=0$ ie $x$ ranges from $-a$ to $a$ (keeping in the mind the anticlockwise rotation) and $x = -(a^2 -y^2)^{\frac12}$ for $x$ ranging from $a$ to $-a$ and carry out the integration I get $\pi a^2$.

But if I do this way $$ 2x dx + 2y dy = 0\quad\text{so}\quad dy = -\frac{x}{y} dx $$ but $y = (a^2-x^2)^(1/2)$ and replacing this for $dy$ in the Integral and keeping rotation anticlockwise I need to find the Integral of $-\frac{x^2}{(a^2- x^2)^(1/2)}$ from $a$ to $-a$, the answer to which comes $-\frac{\pi a^2}{2}$ .

And $y = -((a^2-x^2)^{1/2}$ and replacing this for $dy$ in the Integral and keeping rotation anticlockwise I need to find the Integral of $\frac{x^2}{(a^2- x^2)^{1/2}}$ from $-a$ to $a$, the answer to which comes -$\frac{\pi a^2}{2}$

So the total is $-\pi a^2$

Why is the answer coming $-\pi a^2$ this way when the sense of rotation is not changed. What's the fault ?

$\endgroup$
  • $\begingroup$ Try learning and using MathJax next time you ask a question, to make it more readable: meta.math.stackexchange.com/questions/5020/…. $\endgroup$ – yellon Oct 18 '16 at 18:48
  • $\begingroup$ Observe that $\;x^2+y^2=a^2\implies y=\pm\sqrt{a^2-x^2}\;$ , and for some reason you seem to have ommited the square root in the denominator in the second method. In the answer below I give you another method, the one which, imo, is the easiest one in this case. $\endgroup$ – DonAntonio Oct 18 '16 at 19:06
  • $\begingroup$ Your method is right but what is wrong I'm method. Why is the answer coming -πa^2 ? I had taken the square root. I have corrected it. But what the fault in my method. The answer should have been πa^2 but why is it coming -πa^2 ? $\endgroup$ – user379001 Oct 18 '16 at 19:32
  • $\begingroup$ @user379001 You seem to have done$$\int_{-a}^a\frac{-x^2dx}{\sqrt{a^2-x^2}}=-2\int_0^a\frac{x^2dx}{\sqrt{a^2-x^2}}$$ But the above is wrong: going from $\;-a\;$ to $\;a\;$ on the upper hemisphere gives you negative direction ! The limits must be changed from $\;a\;$ to $\;-a\;$ in the upper hemisphere, and something similar (with changed limits, of course) in thelower hemisphere. That's how you get the correct number with the wrong sign. $\endgroup$ – DonAntonio Oct 18 '16 at 19:47
0
$\begingroup$

Another method which imo is easier than yours, after I commented below your question:

Why not to use the usual, standard parametrization of that circle?:

$$\begin{cases}x=a\cos t\\{}\\y=a\sin t\end{cases}\;\;\;0\le t\le 2\pi\implies dy=a\cos t\,dt$$

and thus the integral is

$$\int_0^{2\pi}a\cos t\cdot a\cos t\,dt=a^2\int_0^{2\pi}\cos^2t\,dt=\left.\frac{a^2}2\left(t+\cos t\sin t\right)\right|_0^{2\pi}=$$

$$=\frac{a^2}2(2\pi)=a^2\pi$$

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Your method is right but what is wrong I'm method. Why is the answer coming -πa^2 ? I had taken the square root. I have corrected it. But what the fault in my method. The answer should have been πa^2 but why is it coming -πa^2 ? $\endgroup$ – user379001 Oct 18 '16 at 19:32
  • $\begingroup$ @user379001 Read my comment below your question. $\endgroup$ – DonAntonio Oct 18 '16 at 19:48
  • $\begingroup$ No for the upper semicircle I have taken y as the positive root and integrated from a to -a ( anticlockwise) and for the lower semicircle , I have taken y as the negative root and integrated from -a to a (anticlockwise ) and that's how I got -πa^2 instead of πa^2 . I have taken different values of y ( the root) , ie. Positive and negative and maintained the anticlockwise rotation but still got -πa^2. Actually if you see the '-' in dy =(-x/y * dx ) is causing the problem but that '-' is but necessary ! $\endgroup$ – user379001 Oct 18 '16 at 20:21
  • $\begingroup$ @user379001 Then you made a mistake in signs somewhere (I can't know and how: you didn't show your actual integration) : I did it your way and it came out correct. $\endgroup$ – DonAntonio Oct 18 '16 at 20:51
  • $\begingroup$ Since I am new to math jax , I can't provide the complete integration but pls check the answer $\endgroup$ – user379001 Oct 18 '16 at 21:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.