1. Find the area of the region bounded by f(x)=x^2 +6x+9 and g(x)=5(x+3). Show the integral used, the limits of integration and how to evaluate the integral.

2. Find the area of the region bounded by x=y^2+6, x=0 , y=-6, and y=7. Show all work required in #1.
3. Find the volume of the solid generated by revolving the region bounded by the graphs of the equations about y=3. Show the integral and give an exact answer.

2 answers

There are 3 questions here.
You show no work , nor do you tell us where your difficulty is.

I will start you off with #1

we need their intersection points
x^2 + 6x + 9 = 5x + 15
x^2 + x - 6 = 0
(x+3)(x-2) = 0
x = -3 or x = 2
if x = -3, y = 0
if x = 2 , y = 25

from x = -3 to x = 2, g(x) > f(x), so the effective height
= 5x + 15 - x^2 - 6x - 9
= 6 - x - x^2

Area = ∫(6 - x - x^2) dx from x = -3 to 2
= [6x - (1/2)x^2 - (1/3)x^3] from -3 to 2
= (12 - (1/2)(4) - (1/3)(8) ) - (-18 - (1/2)(9) - (1/3)(-27) )
= 12 - 2 - 8/3 + 18 - 9/2 + 9
= 179/6
the answer is 125/6 not 179/6 for problem no. 1