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## 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?

- Thread starter sigh1342
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- #1

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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?

- #2

HallsofIvy

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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.## 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

- #3

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

- #4

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

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