Evaluation of a (parabolic) line integral with respect to arc length

[GelatinousFur]

Homework Statement

Evaluate the line integral [tex]\[ \int_c yz\,ds.\][/tex]

where C is a parabola with z=y^2 , x=1 for 0<=y<=2

Homework Equations

A hint was given by the teacher to substitute p=t^2 , dp=(2t)dt and use integration by parts.

I also know from other line integrals with respect to arc length that:

ds=sqrt((dx/dt)^2 + (dy/dt)^2 + (dz/dt)^2)

The Attempt at a Solution

I think that from the information given, the beginning and end points are (1,0,0) to (1,2,4).

My first guess is:
x(t) = t
y(t) = 2t
z(t) = t^2
This will be when t goes from 0 to 2.

So after I have parameterized the curve, I would substitute the functions of t back into the integral to get:

int((2t)^3*sqrt(1^2+2^2+(2t)^2),t,0,2)

=8*int(t^3*sqrt(4t^2+5),t,0,2)

=12032/3

This doesn't look right to me though. Any help would be appreciated!

 

[tiny-tim]

Welcome to PF!

Hi GelatinousFur! Welcome to PF!

C is a parabola with z=y^2 , x=1 for 0<=y<=2
…
My first guess is:
x(t) = t
y(t) = 2t
z(t) = t^2

No, x is constant …

x(t) = 1 …

(and your y and z don't fit each other)

 

[GelatinousFur]

Hi GelatinousFur! Welcome to PF! No, x is constant …

x(t) = 1 …

(and your y and z don't fit each other)

Ah, I see. Thanks for the help (and the welcome)!

So then...

X(t) is constant, so x(t)=1
y(t) =t
z(t) = t^2

when t goes from 0 to 2.

The line integral would then become:

int(y^3,s) over the curve C, because z=y^2.

=int(t^3*sqrt(0^2+1^2+(2t)^2),t,0,2)

=int(t^3*2t,t,0,2)

=2*int(t^4,t,0,2)

=64/5

This answer feels a bit more correct but I still cannot see why the teacher gave us the hint to use integration by parts, as I didn't have to when I just performed that integral.

 

[lanedance]

Ah, I see. Thanks for the help (and the welcome)!

So then...

X(t) is constant, so x(t)=1
y(t) =t
z(t) = t^2

when t goes from 0 to 2.

The line integral would then become:

int(y^3,s) over the curve C, because z=y^2.

=int(t^3*sqrt(0^2+1^2+(2t)^2),t,0,2)

Hi

in you integrand you have
[tex] t^3\sqrt{1+4t^2} [/tex]

I think you simplified away the one from the squareroot

=int(t^3*2t,t,0,2)

=2*int(t^4,t,0,2)

=64/5

This answer feels a bit more correct but I still cannot see why the teacher gave us the hint to use integration by parts, as I didn't have to when I just performed that integral.

 

[GelatinousFur]

Hi

in you integrand you have
[tex] t^3\sqrt{1+4t^2} [/tex]

I think you simplified away the one from the squareroot

Thanks, you are correct.

The correct integral is:

int(t^3*sqrt(1+4*t^2),t,0,2)

So here's where I use integration by parts, but when I integrate sqrt(1+4*t^2) I have to go to an integral table.

I punched int(t^3*sqrt(1+4*t^2),t,0,2) into MATLAB and it spits this answer out:

-1/64/pi^(1/2)*(-3128/15*pi^(1/2)*17^(1/2)-8/15*pi^(1/2))

Is this the wrong answer? Looks a bit weird to me.