Line integral

**Apprentice123** · May 1st 2009, 07:59 AM

1)
$\int_C 3x^2yzds$
$C: x=t, y=t^2, z=\frac{2}{3}t^3 ; (0 \leq t \leq 1)$

My solution

$\int_0^1 2t^7 \sqrt{1+4t^2+4t^4}dt = \frac{13}{20}$

2)
$\int_C (x^2+y^2)dx - xdy$
$C: x^2+y^2 = 1 ; (0 \leq t \leq \frac{ \pi}{2})$

My solution:

$\int_0^{\frac{ \pi}{2}} (-sint - cos^2t)dt$

Is correct ?

**Spec** · May 1st 2009, 08:39 AM

Both are correct. A little trick for the first integral is to use that $4t^4 +4t^2+1=(2t^2+1)^2$

**Apprentice123** · May 1st 2009, 01:06 PM

Thank you

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