# What is the Mistake in the line integral

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 ?

• Try learning and using MathJax next time you ask a question, to make it more readable: meta.math.stackexchange.com/questions/5020/…. – yellon Oct 18 '16 at 18:48
• 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. – DonAntonio Oct 18 '16 at 19:06
• 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 ? – user379001 Oct 18 '16 at 19:32
• @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. – DonAntonio Oct 18 '16 at 19:47

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

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

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

• 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 ? – user379001 Oct 18 '16 at 19:32
• @user379001 Read my comment below your question. – DonAntonio Oct 18 '16 at 19:48
• 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 ! – user379001 Oct 18 '16 at 20:21
• @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. – DonAntonio Oct 18 '16 at 20:51
• Since I am new to math jax , I can't provide the complete integration but pls check the answer – user379001 Oct 18 '16 at 21:19