Simple Line Integral becomes troublesome

gipc
Messages
69
Reaction score
0
I have the parametrization of C
x=9cos(t)
y=9sin(t)
z=-8t

0<=t<=10*pi

and I have to calculate \int_c 7x^2+4y^2 -5xy

after I transform this I get
(7*81cos^2(t) +4*81sin^2(t) -5*81*cos(t)*sin(t))*sqrt(-9sin^2(t)+9cos^2(t)-8) dt from 0 to 10*pi

Now that is a monster.
What did I do wrong?
 
Physics news on Phys.org
Obtain the values of x, y, z in terms of the boundary conditions for t, i.e. t = 0 and t = 10\pi. It should be easier to solve.
 
When t= 0 x= 9, y= 0, z= 0.

When t= 10\pi, x= 9, y= 0, z= -80\pi.

I don't see how that helps at all!

gipc, you statement \int_c 7x^2+ 4y^2- 5z^2 is missing the differential. Is that with respect to dx, dy, dz, or the arclength ds, or what?

Because of the square root, I assume it is ds but it would not be what you give. it would be
\sqrt{81 sin^2(t)+ 81 cos^2(t)+ 64}dt= \sqrt{81+ 64}dt= \sqrt{145}dt

Did you forget that sin^2(t)+ cos^2(t)= 1?
 
yes, i did forget :)
thanks for that, that was the mistake that was holding me back
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top