The region is a right circular cylinder of radius 3 , with the bottom at −7 and top at 7 . find the limits of integration on the triple integral for the volume of the cylinder using cartesian, cylindrical, and spherical coordinates and the function to be integrated. for your answers θ= theta, ϕ= phi, and rho= rho. cartesianv=∫ba∫dc∫fep(x, y,z) dydxdzwhere a= , b= , c= , d= , e= , f= and p(x, y,z)= .cylindricalv=∫ba∫dc∫fep(r,θ,z)drdθ dzwhere a= , b= , c= , d= , e= , f= and p(r,θ,z)= .spherical (which is twice the top half)hint: here you must write the volume as the sum of two integrals 0≤ϕ≤arctan(3/7) and arctan(3/7)≤ϕ≤π/2v=2[∫b1a1∫d1c1∫f1e 1p(rho,θ,ϕ)drhodϕdθ+∫b2a2∫d2c2∫f2e2 p(rho,θ,ϕ)drhodϕdθ]where a1= , b1= , c1= , d1= , e1= , f1= and p(rho,θ,ϕ)= .andwhere a2= , b2= , c2= , d2= , e2= , f2= and p(rho,θ,ϕ)=

"image"

step-by-step explanation:

in most of these problems, the notation is that the point with a "prime" is the image of the point without. that is, b' is the image of b.

here, there is no dilation or reflection. there is simply translation (with no direction shown). which figure you call the "preimage" and which one you call the "image" is up to you. as a matter of convention, the naming is as described above.

central angle of a triangle has the same measure as its intercepted arc.

step-by-step explanation:

step-by-step explanation:

given that ab is divided into equal parts as

ac, cd, de, ef, fg, gh, hi, ib.

from this we find that f is the midpoint of ab.

so using mid point formula we find

f = (3,7)

similarly considering fb, we get

h = mid point of fb = ((6,6) and g, mid point of fh is

g = (9/2, 13/2)

i is mid point of hb

so i =(15/2,11/2)