Can you show us your calculation in full?
Calc3 multivariable cylindrical coordinates!Integrate the function f(xyz)=−6x+5y over the solid given by the figure below, if P=(-6,-6,3), Q=(-3,-3,3), R=(3,3,3), and S=(6,6,3). (The figure is based on the xy-plane.) the figure seems to be touching or on the xy plane.
i don't know how to paste the picture but it's basically taking the bigger cylinder and subtracting it from the small cylinder to get a thin hollowed cylinder
so i know that in order to find the range for the angles you need to take the tan() of the x and y: tan(theta)=-6/6, theta=5pi/4 tan(theta)=3/3 theta=pi/4
the radius is just sqrt(3^2 + 3^2)=sqrt(18) and sqrt(6^2 + 6^2)=sqrt(72)
so my triple integral would be:
integral(theta=pi/4 to theta=5pi/4) of integral(r=sqrt(18) to r=sqrt(72)) of integra(0 to 3) of -6x+5y...
i've tried 5 times and i got the answer wrong. I need help!!
-6x+5y in cylindrical is -6rcos(theta)+5rsin(theta)*rdzdrd(theta)
let's set -6rcos(theta)+5rsin(theta)=A
so first integral is dz:
integral(from z=0 to z=3) of A dz = z(-6rcos(theta)+5rsin(theta)) + C |z=0 to z=3
so that should result in -18r^2cos(theta)+15r^2sin(theta) (we'll call this answer B)
the next integral is r:
integral(from r=sqrt(18) to sqrt(72)) of B = -6r^3cos(theta)+5r^3sin(theta)+c |r=sqrt(18) to r=sqrt(72)
so that should result in -6(sqrt72)^3cos(theta)+5(sqrt72)^3sin(theta) + 6(sqrt18)^3cos(theta)-5(sqrt18)^3sin(theta)
and then you take the last integral of theta which shouldn't be too hard...
and you plug in pi/4 to 5pi/4...i keep getting 8316. but that's what too huge of an answer or at least that is the wrong answer anyways/