Let the region R be the area enclosed by the function f(x)=x^2+2f(x)=x 2
+2, the horizontal line y=8,y=8, and the yy-axis. If the region R is the base of a solid such that each cross section perpendicular to the xx-axis is a semi-circle with diameters extending through the region R, find the volume of the solid. You may use a calculator and round to the nearest thousandth.

there are lots of ways.   pick's theorem tells us if the vertices are lattice points the area a is

a = b/2 + i - 1

where b is the number of boundary lattice points and i is the number of interior lattice points.

in the figure i count b=16, i=50 so

a = 8 + 50 - 1 = 57

let's try the shoelace formula.   the area is half the sum of the cross products of the sides.  

a(-2,6)   -2(2)-6(-5) = 26

b(-5,2)   -5(-3) - 2(-2) = 19

c(-2,-3) -2(-3) - (-3)(4) = 18

d(4,-3)   4(3) - (-3)(5) = 27

e(5,3)   5(3) - 3(1) = 12

f(1,3)     1(6) - 3(-2) = 12

a(-2,6)

each of the cross products is twice the area of the triangle with the associated side and one vertex the origin. so the area of abo=26/2=13.

area = (1/2) (26 + 19 +   18 + 27 + 12 + 12) = 57

that checks!

i think 5 miles per hour

step-by-step explanation:

answer would be 8.64 or s = 11π/4 inches

step-by-step explanation:

s = r θ  

where s = arc length  

r = radius = 3 inches  

θ = central angle in radians = ( π /180)*165  

s = 3(π/180)*165  

= 8.64 inches