Path integral（Parametric curve）

[sigh1342]

Homework Statement

Find $$\int_{C} z^3 ds $$ where C is the part of the curve $$ x^2+y^2+z^2=1,x+y=1$$ where$$ z ≥ 0 $$ then I let $$ x=t , y=1-t , z= \sqrt{2t-2t^2}$$ . Is it correct? Or there are some better idea?

Homework Equations

The Attempt at a Solution

 

[HallsofIvy]

Homework Statement

Find $$\int_{C} z^3 ds $$ where C is the part of the curve $$ x^2+y^2+z^2=1,x+y=1$$ where$$ z ≥ 0 $$ then I let $$ x=t , y=1-t , z= \sqrt{2t-2t^2}$$ . Is it correct? Or there are some better idea?

Yes, that is correct. And, of course, $ds= \sqrt{dx^2+ dy^2+ dz^2}= \sqrt{1+ 1+ (2- 2t)^2/(2t- 2t^2)} dt$. I thought about using the standard parameterization of the sphere and then adding the condition that x+y= 1, in order to avoid the square root, but that does not appear to give a simpler integral.

Homework Equations

The Attempt at a Solution

 

[cosmic dust]

This is OK, just note that 0$\leq$t$\leq$1 . The answer sjould be 1/3.

 

[clamtrox]

Yes, that is correct. And, of course, $ds= \sqrt{dx^2+ dy^2+ dz^2}= \sqrt{1+ 1+ (2- 2t)^2/(2t- 2t^2)} dt$. I thought about using the standard parameterization of the sphere and then adding the condition that x+y= 1, in order to avoid the square root, but that does not appear to give a simpler integral.

There is actually a small typo in your ds, which makes it much more complicated than it actually is.