The discussion revolves around evaluating the integral $$\int_{C} z^3 ds $$ along a specified curve defined by the equations $$ x^2+y^2+z^2=1 $$ and $$ x+y=1 $$, with the condition that $$ z \geq 0 $$.
The discussion includes confirmations of the parameterization and calculations related to the differential element $$ ds $$, with some participants noting potential complications in the expression for $$ ds $$ and suggesting that it may be more complex than necessary. There is no explicit consensus on the best approach, but multiple lines of reasoning are being explored.
Participants note the constraint $$ 0 \leq t \leq 1 $$ and mention a possible typo in the expression for $$ ds $$ that could complicate the calculations.
Yes, that is correct. And, of course, [itex]ds= \sqrt{dx^2+ dy^2+ dz^2}= \sqrt{1+ 1+ (2- 2t)^2/(2t- 2t^2)} dt[/itex]. 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.sigh1342 said: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 said:Yes, that is correct. And, of course, [itex]ds= \sqrt{dx^2+ dy^2+ dz^2}= \sqrt{1+ 1+ (2- 2t)^2/(2t- 2t^2)} dt[/itex]. 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.