Question

hey is there anyone who can help me with converting a triple integral to cylindrical coordinates?? cheers!

Answer 1

particularly with the example...:
I=int[-1 to 1] int[-sqrt(1-x^2) to sqrt(1-x^2)] int[0 to 2*sqrt(x^2 + y^2)] x^2dzdydx

Answer 2

answer is given as \[\int\limits_{0}^{2\Pi}\int\limits_{0}^{1}\int\limits_{0}^{2r}r^2\cos^2\Theta rdzdrd \Theta\]
and I understand everything apart from how to get 0 to 2Pi for the theta coordinates....
Help please! :D

Answer 3

Sorry - original equation is \[\int\limits_{-1}^{1}\int\limits_{-\sqrt(1-x^2)}^{\sqrt(1-x^2}\int\limits_{0}^{2\sqrt(x^2+y^2)}x^2dzdydx\]

Answer 4

Feldy, it's a circle. The limits of your dy part of your integral are y = sqrt(1-x^2) and y = - sqrt(1-x^2). If you square both sides on each of these equations, you will get the equation of a circle with radius one. If you draw the cricle, draw a line vertically simple since dy is first, it touches the top and the bottom of the circle. Therefore, the region of intrest spans the entire circle....from 0 to 2 pi.

Answer 5

your volume is the area between a cone and the xy plane over the unit disc.

Answer 6

x^2+y^2 = r^2 for the polar substution

Answer 7

x= r cos(theta)

Answer 8

and the jacobian of the substitution is r

Answer 9

the limits on x and y give a unit disc which in polar coordinates is 0<=theta<=2pi

and 0<=r<=1

Answer 10

Cheers guys... so Nisha, if , say my y limits were 0 to y=sqrt(1-x^2), instead of -sqrt(1-x^2), then it would be 0 to Pi, instead of 2Pi???

Answer 11

yes

Answer 12

would only trace out the first and second quadrants

Answer 13

yes

Answer 14

Thank you so much!