# Math Help - Line integral

1. ## Line integral

Evaluate the line integral $\int_C(x+y^2)$, where $C$ is the parabola $y=x^2$ in the plane $z=0$ connecting the points $(0,0,0)$ and $(2,4,0)$.

My method:

$\gamma (t)=(2t, 4t^2,0)$ where $t \in [0,1]$.

I need to find $\int_0^1 f(\gamma (t)) || \gamma (t) ||~dt$ where $f(x,y)=x+y^2$

$\int_0^1 (2t+(4t^2)^2)(2^2+(8t^2))^{\frac{1}{2}}~dt$

$\int_0^1 (2t+16t^4)(4+64t^2)^{\frac{1}{2}}~dt$

$2^2 \int_0^1 (t+8t^4)(1+16t^2)^{\frac{1}{2}}~dt$

$4 \int_0^1 t \sqrt{1+16t^2}~dt+32 \int_0^1 t^4 \sqrt{1+16t^2}~dt$

For the first integral:

Let $u=1+16t^2 \Rightarrow \ \frac{du}{dt}=32t$

$4 \int_1^{17} \sqrt{u} ~\frac{du}{32} \Rightarrow \frac{1}{8}\int_1^{17}\sqrt{u}~du$

$\frac{1}{8} \left[ \frac{3}{2} u^{\frac{2}{3}} \right]_1^{17}$

$=\frac{3}{16}(17^{\frac{2}{3}}-1)$

So the two integrals are now:

$\frac{3}{16}(17^{\frac{2}{3}}-1)+32 \int_0^1 t^4 \sqrt{1+16t^2}~dt$

For the second integral:

Let $(4t)^2=\cosh 2u$

Then, for the second integral, we have $32 \int_{\frac{1}{2}arcosh 0}^{\frac{1}{2} arcosh 16} \frac{1}{16} (\cosh^2 2u)( \cosh 2u) \frac{1}{4} \left( \frac{1}{2} \right) (\cosh 2u)^{\frac{1}{2}}(2 \sinh 2u)~du$

Which becomes:

$\frac{1}{2} \int_{\frac{1}{2} arcosh 0}^{\frac{1}{2} arcosh 16}(\cosh 2u)^{\frac{7}{2}} \sinh 2u~du$

by FTC, the integral is $\left[ \frac{1}{18} (\cosh 2u)^{\frac{9}{2}} \right]_{\frac{1}{2} arcosh 0}^{\frac{1}{2} arcosh 16}$

so substituting the limits gives:

$\frac{1}{18} (16)^{\frac{9}{2}}$

So my overall answer to the question is:

$\frac{3}{16}(17^{\frac{2}{3}}-1)+\frac{1}{18} (16)^{\frac{9}{2}}$

Firstly, is this right? I have an $arcosh 0$ in there that is undefined, but it removes itself quite nicely at the end.

Secondly, is there a faster way of doing this???

2. Originally Posted by Showcase_22
Evaluate the line integral $\int_C(x+y^2)$, where $C$ is the parabola $y=x^2$ in the plane $z=0$ connecting the points $(0,0,0)$ and $(2,4,0)$.

My method:

$\gamma (t)=(2t, 4t^2,0)$ where $t \in [0,1]$.

I need to find $\int_0^1 f(\gamma (t)) || \gamma (t) ||~dt$ where $f(x,y)=x+y^2$

== Probably a typo: if must be $|| \gamma^{'} (t)||$ in the integral

$\int_0^1 (2t+(4t^2)^2)(2^2+(8t^2))^{\frac{1}{2}}~dt$

$\int_0^1 (2t+16t^4)(4+64t^2)^{\frac{1}{2}}~dt$

$2^2 \int_0^1 (t+8t^4)(1+16t^2)^{\frac{1}{2}}~dt$

$4 \int_0^1 t \sqrt{1+16t^2}~dt+32 \int_0^1 t^4 \sqrt{1+16t^2}~dt$

For the first integral:

Let $u=1+16t^2 \Rightarrow \ \frac{du}{dt}=32t$

$4 \int_1^{17} \sqrt{u} ~\frac{du}{32} \Rightarrow \frac{1}{8}\int_1^{17}\sqrt{u}~du$

$\frac{1}{8} \left[ \frac{3}{2} u^{\frac{2}{3}} \right]_1^{17}$

$=\frac{3}{16}(17^{\frac{2}{3}}-1)$

=== This is wrong: $\int{\sqrt{u} \, du} = \frac{2}{3}u^{\frac{3}{2}}$ and not what you wrote

Tonio

So the two integrals are now:

$\frac{3}{16}(17^{\frac{2}{3}}-1)+32 \int_0^1 t^4 \sqrt{1+16t^2}~dt$

For the second integral:

Let $(4t)^2=\cosh 2u$

Then, for the second integral, we have $32 \int_{\frac{1}{2}arcosh 0}^{\frac{1}{2} arcosh 16} \frac{1}{16} (\cosh^2 2u)( \cosh 2u) \frac{1}{4} \left( \frac{1}{2} \right) (\cosh 2u)^{\frac{1}{2}}(2 \sinh 2u)~du$

Which becomes:

$\frac{1}{2} \int_{\frac{1}{2} arcosh 0}^{\frac{1}{2} arcosh 16}(\cosh 2u)^{\frac{7}{2}} \sinh 2u~du$

by FTC, the integral is $\left[ \frac{1}{18} (\cosh 2u)^{\frac{9}{2}} \right]_{\frac{1}{2} arcosh 0}^{\frac{1}{2} arcosh 16}$

so substituting the limits gives:

$\frac{1}{18} (16)^{\frac{9}{2}}$

So my overall answer to the question is:

$\frac{3}{16}(17^{\frac{2}{3}}-1)+\frac{1}{18} (16)^{\frac{9}{2}}$

Firstly, is this right? I have an $arcosh 0$ in there that is undefined, but it removes itself quite nicely at the end.

Secondly, is there a faster way of doing this???

I inserted a couple of times in your text above

Tonio

3. Originally Posted by Showcase_22
Evaluate the line integral $\int_C(x+y^2)$, where $C$ is the parabola $y=x^2$ in the plane $z=0$ connecting the points $(0,0,0)$ and $(2,4,0)$.

My method:

$\gamma (t)=(2t, 4t^2,0)$ where $t \in [0,1]$.

I need to find $\int_0^1 f(\gamma (t)) || \gamma \ ' (t) ||~dt$ where $f(x,y)=x+y^2$

$\int_0^1 (2t+(4t^2)^2)(2^2+(8t^2))^{\frac{1}{2}}~dt$

$\int_0^1 (2t+16t^4)(4+64t^2)^{\frac{1}{2}}~dt$

$2^2 \int_0^1 (t+8t^4)(1+16t^2)^{\frac{1}{2}}~dt$

$4 \int_0^1 t \sqrt{1+16t^2}~dt+32 \int_0^1 t^4 \sqrt{1+16t^2}~dt$

For the first integral:

Let $u=1+16t^2 \Rightarrow \ \frac{du}{dt}=32t$

$4 \int_1^{17} \sqrt{u} ~\frac{du}{32} \Rightarrow \frac{1}{8}\int_1^{17}\sqrt{u}~du$

$\frac{1}{8} \left[ \frac{2}{3} u^{\frac{2}{3}} \right]_1^{17}$

$=\frac{1}{12}(17^{\frac{2}{3}}-1)$

So the two integrals are now:

$\frac{1}{12}(17^{\frac{2}{3}}-1)+32 \int_0^1 t^4 \sqrt{1+16t^2}~dt$

For the second integral:

Let $(4t)^2=\cosh 2u$

Then, for the second integral, we have $32 \int_{\frac{1}{2}arcosh 0}^{\frac{1}{2} arcosh 16} \frac{1}{16} (\cosh^2 2u)( \cosh 2u) \frac{1}{4} \left( \frac{1}{2} \right) (\cosh 2u)^{\frac{1}{2}}(2 \sinh 2u)~du$

Which becomes:

$\frac{1}{2} \int_{\frac{1}{2} arcosh 0}^{\frac{1}{2} arcosh 16}(\cosh 2u)^{\frac{7}{2}} \sinh 2u~du$

by FTC, the integral is $\left[ \frac{1}{18} (\cosh 2u)^{\frac{9}{2}} \right]_{\frac{1}{2} arcosh 0}^{\frac{1}{2} arcosh 16}$

so substituting the limits gives:

$\frac{1}{18} (16)^{\frac{9}{2}}$

So my overall answer to the question is:

$\frac{1}{12}(17^{\frac{2}{3}}-1)+\frac{1}{18} (16)^{\frac{9}{2}}$

Firstly, is this right? I have an $arcosh 0$ in there that is undefined, but it removes itself quite nicely at the end.

Secondly, is there a faster way of doing this???
eep! Both of those were typos! I remember seeing the second mistake you pointed out and thinking "gotta change that".

*facepalm*

But, overall, is this right now?

4. Originally Posted by Showcase_22
eep! Both of those were typos! I remember seeing the second mistake you pointed out and thinking "gotta change that".

*facepalm*

But, overall, is this right now?

No, it isn't imo: you haven't yet changed the power of 17 in your answer!

Why don't you do it the simple way: we're gonna go on $y=x^2$ from $\left(0,0,0\right)$ to $\left(2,4,0\right)$ , so why don't you choose $x$ as parameter, with $0\leq x \leq 2$??

This way you have to evaluate the integral

$\int_0^2\left(x+x^4\right)\sqrt{\left(1+2x^2\right )}~dx$

Tonio

5. Originally Posted by Showcase_22
Evaluate the line integral $\int_C(x+y^2)$, where $C$ is the parabola $y=x^2$ in the plane $z=0$ connecting the points $(0,0,0)$ and $(2,4,0)$.

My method:

$\gamma (t)=(2t, 4t^2,0)$ where $t \in [0,1]$.

I need to find $\int_0^1 f(\gamma (t)) || \gamma \ ' (t) ||~dt$ where $f(x,y)=x+y^2$

$\int_0^1 (2t+(4t^2)^2)(2^2+(8t)^2)^{\frac{1}{2}}~dt$

$\int_0^1 (2t+16t^4)(4+64t^2)^{\frac{1}{2}}~dt$

$2^2 \int_0^1 (t+8t^4)(1+16t^2)^{\frac{1}{2}}~dt$

$4 \int_0^1 t \sqrt{1+16t^2}~dt+32 \int_0^1 t^4 \sqrt{1+16t^2}~dt$

For the first integral:

Let $u=1+16t^2 \Rightarrow \ \frac{du}{dt}=32t$

$4 \int_1^{17} \sqrt{u} ~\frac{du}{32} \Rightarrow \frac{1}{8}\int_1^{17}\sqrt{u}~du$

$\frac{1}{8} \left[ \frac{2}{3} u^{\frac{3}{2}} \right]_1^{17}$

$=\frac{1}{12}(17^{\frac{3}{2}}-1)$

So the two integrals are now:

$\frac{1}{12}(17^{\frac{3}{2}}-1)+32 \int_0^1 t^4 \sqrt{1+16t^2}~dt$

For the second integral:

Let $(4t)^2=\cosh 2u$

Then, for the second integral, we have $32 \int_{\frac{1}{2}arcosh 0}^{\frac{1}{2} arcosh 16} \frac{1}{16} (\cosh^2 2u)( \cosh 2u) \frac{1}{4} \left( \frac{1}{2} \right) (\cosh 2u)^{\frac{1}{2}}(2 \sinh 2u)~du$

Which becomes:

$\frac{1}{2} \int_{\frac{1}{2} arcosh 0}^{\frac{1}{2} arcosh 16}(\cosh 2u)^{\frac{7}{2}} \sinh 2u~du$

by FTC, the integral is $\left[ \frac{1}{18} (\cosh 2u)^{\frac{9}{2}} \right]_{\frac{1}{2} arcosh 0}^{\frac{1}{2} arcosh 16}$

so substituting the limits gives:

$\frac{1}{18} (16)^{\frac{9}{2}}$

So my overall answer to the question is:

$\frac{1}{12}(17^{\frac{3}{2}}-1)+\frac{1}{18} (16)^{\frac{9}{2}}$

Firstly, is this right? I have an $arcosh 0$ in there that is undefined, but it removes itself quite nicely at the end.

Secondly, is there a faster way of doing this???
Right, it's been amended.

I don't think it's much easier parameterising between 0 and 2 since my substitution is now $(\sqrt{2}x)^2= \cosh 2u$ which isn't that bad, but it is slightly more difficult than $(4x)^2=\cosh 2u$.