# Very basic line integrals (worked out)

my problem is basically (among many others) my textbook sucks and the answers are frequently non nonsensical, sadly i haven't slept in a couple days so im not sure whos out to lunch. gonna post the problem my steps to the answer and the answer in my book if you don't mind pointing out my mistake id really appreciate it.

(i) Find the arc length of the parametrized curve.

Question: $g(t)=(\frac{t^{3}}{3}-t,t^{2})$ where $t \in [0,2]$

Arc-length = $\int_{C}ds$

where $ds= \sqrt {(t^{2}-1)^{2}+(2t)^{2}}$

$ds= \sqrt {(t^{2}+1)^{2}}=t^{2}+1$

$\int^{2}_{0}t^{2}+1 =(\frac{t^{3}}{3}+t) |^{2}_{0}=\frac{14}{3}$ my book has the answer as 4.

(ii) Compute $\int_{C} \sqrt z ds$ where C is paramentrized by $g(t)=(2\cos t ,2\sin t , t^{2})$ where $t\in [0,2\pi ]$

$ds= \sqrt {(-2\sin t)^{2}+(2\cos t)^{2}+(2t)^{2}}$

$ds= \sqrt {4+4t^{2}}=2\sqrt {1+t^{2}}$

$\int^{2\pi}_{0} \sqrt{t^{2}}2\sqrt {1+t^{2}}dt$=$\int_{l} \sqrt {u}du$=$\frac {2u^{3/2}}{3}$= $\frac {2}{3}((1+t^{2})^{\frac {3}{2}})|^{2\pi}_{0}$

=$\frac {2}{3}(1+4\pi^{2})^{\frac {3}{2}}$ my text books answer is $\frac {1}{3}(1+4\pi^{2})^{\frac {3}{2}}$ so at least were close this time O.o

(iii) Compute $\int_{C}$ F $\cdot$ d x

F(x,y)=$(x^{2}y,x^{3}y^{2})$

C is the closed curve formed by portions of the line y=4 and $y=x^{2}$ orientated counterclockwise.

In this case i need to parametrize the function myself i decide to pick $x=t$ thus it follows $y=t^{2}$

so d x = $(1,2t)dt$ and F(t)=$(t^{2}t^{2},t^{3}t^{4})$ thus

F(t) $\cdot$ d x =$(t^{4},t^{7})\cdot (1,2t)dt$ = $t^{4}+2t^{8}$

$\int_{C}t^{4}+2t^{8}$=$(\frac {t^{5}}{5}+\frac {2t^{9}}{9})|_{C}$

Now here i actually run into quite a bit of problems what the hell C? Logically x is moving from -$2 \to 2$ and y is moving from $0\to4$.

But clearly as t goes from $0\to2$, I cover all of $y$ but somethings not right with x did i do my parametrization incorrectly?

Basically what i have is $\frac{9t^{5}+10t^{9}}{45}$ when i evaluate t $\in [0,2]$ i get $\frac{9(32)+10(512)}{45}$ =$\frac{5408}{45}$

my textbooks answer is $\frac{9856}{45}$ since 7|9856 with r=0 i have a very big problem if this is correct as $3^{n}2^{l}+5^{k}2^{m}$ where $(n,l,k,m) >0$ and are integers is never divisible by 7 which would mean my answer is clearly out to lunch. being that my denominator is correct i am probably out by some constant in my parametrization.

I realize this is a rather large question but i think it will be fairly little to no effort involved to answer.

Cheers and thank you for the help!

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why in (i) is $\frac{d}{dt} t^3 = t^2 -1$? – pppqqq Apr 3 '13 at 11:10
fixed now, parametrization wasn't written down correctly – Faust7 Apr 3 '13 at 11:49
Ok... in (i) i'm getting your sam result, in (iii) you did the integration only in the first half of curve! – pppqqq Apr 3 '13 at 11:59
im sorry can you please explain? i havent slept in a couple days you'll need to a bit more pointed for me to follow – Faust7 Apr 3 '13 at 12:09
If I understand well the question, you have to evalaute the work over the path which is A) $y=x^2, x\in (-2,2)$ plus B) \$y=4 x\in (2,-2)... did you do B)? – pppqqq Apr 3 '13 at 12:15