Solving a Line Integral Problem: Struggling for the Right Answer

galaxy_twirl
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Hi everyone. I ran into a minor problem while trying to solve a problem on line integral. I suspect this question to be very straight-forward as it gave the parametric equations of the curve C. However, I am still unable to get the answer for some reason. May I have someone to point out where I went wrong? Thanks!

Question:

Ah, the image didn't get uploaded.. Please refer to the attached image. Thanks! The question in question is Q1(a).

My attempt:

Since x = t2 + 1 and y = t3+t , we have x(t) and y(t). Following on, I went one to apply the formula, ∫ F⋅ dr = ∫ f(x(t), y(t))||r'(t)|| dt.

I had r(t) as (t2 + 1) i + (t3+t) j. And I have r'(t) = (2t) i + (3t2 + 1) j, and ||r'(t)|| = √ (9t2 + 1)(t2 + 1).

Since the range of t is 0 ≤ t ≤ 1, I integrated over this range and my calculator returned 3.73 as the answer, but the answer in my answer key wrote 6. May I know where I had gone wrong? Thanks! :D

<In case the picture cannot be seen, I have uploaded it here.> Thanks again!
 

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galaxy_twirl said:
Hi everyone. I ran into a minor problem while trying to solve a problem on line integral. I suspect this question to be very straight-forward as it gave the parametric equations of the curve C. However, I am still unable to get the answer for some reason. May I have someone to point out where I went wrong? Thanks!

Question:

Ah, the image didn't get uploaded.. Please refer to the attached image. Thanks! The question in question is Q1(a).

My attempt:

Since x = t2 + 1 and y = t3+t , we have x(t) and y(t). Following on, I went one to apply the formula, ∫ F⋅ dr = ∫ f(x(t), y(t))||r'(t)|| dt.

I had r(t) as (t2 + 1) i + (t3+t) j. And I have r'(t) = (2t) i + (3t2 + 1) j, and ||r'(t)|| = √ (9t2 + 1)(t2 + 1).
Your r'(t) looks fine, but |r'(t)| doesn't.
galaxy_twirl said:
Since the range of t is 0 ≤ t ≤ 1, I integrated over this range and my calculator returned 3.73 as the answer, but the answer in my answer key wrote 6. May I know where I had gone wrong? Thanks! :D

<In case the picture cannot be seen, I have uploaded it here.> Thanks again!
 
Can you show me your work on the finding the gradient of f?
 
To Mark44: Oh dear. I must have been hallucinating when doing Math at nearly 3am. T_T Sorry. ||r'(t)|| is √(t^2 + (1/9))(t^2 + 1).

To td21: How do I find the gradient of f? Do I just use the formula Gradient = (y2-y1)/(x2-x1)?
 
However, my calculator returned 1.24 as the answer.. :/
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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