Convert the integrals into polar coordinate form. x is from 0 to 2, y is o to sqrt(2x-x^2)?

Question

amistre64 · Answer

r=sqrt(x^2+y^2)

amistre64 · Answer

and theta = tan-1(y/x)

amistre64 · Answer

r = sqrt(x^2+2x-x^2) = sqrt(2x)

anonymous · Answer

what does tan-1(y/x) mean?

amistre64 · Answer

arctangent function

amistre64 · Answer

tan(t) = y/x

t = tan-1(y/x)

anonymous · Answer

The question ask to integrate those over 5*sqrt(x^2+y^2) though

anonymous · Answer

Those were the integrals.

amistre64 · Answer

amistre64 · Answer

most likely you have a vector valued function and want to integrate it

anonymous · Answer

Question is "Evaluate the iterated integral by converting to polar coordinates."

anonymous · Answer

the function is 5*sqrt(x^2+y^2), and the integrals are 0 to 2 for x, 0 to sqrt(2x-x^2) for y, dy dx.

amistre64 · Answer

sounds vaguely familiar

amistre64 · Answer

$$\int_{0}^{2}\int_{0}^{2x-x^2}5\sqrt{x^2+y^2}\ dydx$$

anonymous · Answer

2x-x^2 is square-rooted

amistre64 · Answer

r = sqrt(x^2+y^2 sooo)  5r 

$$\int_{0}^{2}\int_{0}^{\sqrt{2x-x^2}}5r\ dydx$$
and know your asking to change the limits ....

amistre64 · Answer

i believe r = sqrt(2x) is still valid for dr

amistre64 · Answer

oy!!  5 sqrt(x^2+y^2 ) is a halfcircle centered at the origin with a radius of 5

anonymous · Answer

.....And then what? :o

anonymous · Answer

integral would be 0 to pi/2 and 0 to 5?

anonymous · Answer

But the answer doesn't have any pi, which is very weird!

amistre64 · Answer

amistre64 · Answer

helps if i can see these things right :)

anonymous · Answer

Sorry dude but I really don't get it. But thanks for trying to get me to understand it

amistre64 · Answer

'sok, still kinda fuzzy on these myself

amistre64 · Answer

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this is a rough idea of our domain

amistre64 · Answer

and "above" this is our surface 5sqrt(x^2+y^2)

anonymous · Answer

Yup i got that part actually. i just cant figure out the integrals! :/

amistre64 · Answer

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