Question

change the cartesian integral into polar integral:
integral -a to a, integral -(a^2-x^2)^1/2 to (a^2-x^2)^1/2 dy dx

Answer 1

all you have to do is substitute x=rCos(theta) and y=rSin(theta) and change your integral to r*dr*dtheta

Answer 2

tried that

Answer 3

i got 1/2 a^2- 1/2a^2

Answer 4

r=+-a

Answer 5

what is your function? y=(a^2-x^2)^1/2 or is this equal to a constant?

Answer 6

y^2=a^2-x^2
y^2+x^2=a&2
R^2=a^2
r=+/- a

Answer 7

so my first integral becomes -a to a of r dr? correct?

Answer 8

so my first integral becomes -a to a of r dr? correct?

Answer 9

so this is just a circle? with radius a? if so then you have int(0,2*pi)int(0,a)r*dr*dtheta

Answer 10

so this is just a circle? with radius a? if so then you have int(0,2*pi)int(0,a)r*dr*dtheta

Answer 11

OMG!! r=a is a circle from 0 to a, you are right! I was thinking it was from -a to a which is a diameter not a radius, thank you!!!

Answer 12

not a problem, anytime