Set up the iterated integral for evaluating modifyingbelow integral from nothing to nothing integral from nothing to nothing integral from nothing to nothing with upper d f (r comma theta comma z )dz r dr d theta∫∫∫df(r,θ,z)dz r dr dθ over the region d, where d is the solid right cylinder whose base is a region in the xy-plane that lies inside the cardioid r equals 2 plus 2 cosine thetar=2+2cosθ and outside the circle r equals 2r=2, and whose top lies in the plane z equals 6z=6.

answer: hi

step-by-step explanation:

step-by-step explanation:

here's what happens:

1) the graph of f   = x^2 is shifted 2 units to the right.

2) the resulting graph is reflected in the x-axis, so that the parabola now opens down.

3) this final result is transformed up by 5 units.

a function could be identified in 3 ways either on a chart,graph, or ordered pairs. first a function is that each point has its own. the rule is that there can be many x's to one y but there can not be 2 x's at different y's. so on a graph when you put a straight line through the y's there should be one dot only to be a function.

step-by-step explanation: