I review (pretest drill) which i posted and no one help me. If you don't want to do it because you can't write the Latax in here. You also didn't help me with any clues. So sad...
A. consider the region R in the first quarant that is outside the circle r=1 and inside the four-leaved rose r = 2sin2theta.
a) draw sketch of the circle and four-leaved rose and the shade the region R.
b) write the following double integral as an iterated in polar coordinates. Do not evaluate.
double integral of cos 2thetadA (see picture attached)
c). evaluate the integral...i can do this part.
B. consider the following iterated integral
Double integral of ydxdy ( limit respect to y from -4 to 0, limit integral respect to x from -sqrt(16-y^2) to sqrt(16-y^2)
b1. sketch the region of integration
b2. rewrite the integral in order dydx
your help will distribute a lot of to my math skills. Thanks for your help....!
b3. rewrite the integral in polar coordinates
b4. evaluate
I assume you know how to sketch figures in polar coordinates.
I will represent (theta) as Ob) write the following double integral as an iterated in polar coordinates. Do not evaluate.
double integral of cos 2thetadA (see picture attached)
INT cos(2O) dA = Int Int cos(2O)*r dr dO
The domain of this integration is of the bottom half of a circle (if I'm not mistaken). The radius of this circle is 4 and the center is at the origin. The function that we are integrating, f(x,y) = y is a flat plane in 3D space that bisects the y-z axis. You'll have to do your best to try and draw this because I cannot do a 3D graph on this site (that I know of).c). evaluate the integral...i can do this part.
B. consider the following iterated integral
Double integral of ydxdy ( limit respect to y from -4 to 0, limit integral respect to x from -sqrt(16-y^2) to sqrt(16-y^2)
b1. sketch the region of integration
b2. rewrite the integral in order dydx
The integral can be rewritten in the order dy dx by changing the domain of integration to D = {(x,y) | -4 <= x <= 4, -sqrt(16 - x^2) <= y <= 0}
Int{x = -4, 4} INT{y = -sqrt(16 - x^2), 0} y dy dx
(I could be mistaken on this. It's been a while since I've done a problem like this.)
D = {(r,O) | 0 <= r <= 4, -pi <= O <= 0}your help will distribute a lot of to my math skills. Thanks for your help....!
b3. rewrite the integral in polar coordinates
b4. evaluate
dy dx = dA = r dr dO
rsinO = y
rcosO = x
Int{O = -pi, 0} Int{r = 0, 4} rsinO*r dr dO
= Int{O = -pi, 0} Int{r = 0, 4} r^2*sinO dr dO