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.

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

1. To find the area of the region bounded by the curves f(x) = x^2 + 6x + 9 and g(x) = 5(x + 3), we need to find the points where they intersect and set up the integral accordingly.

First, let's find the points of intersection:
f(x) = g(x)
x^2 + 6x + 9 = 5(x + 3)
x^2 + 6x + 9 = 5x + 15
x^2 + x - 6 = 0

Factoring this quadratic equation, we get:
(x + 3)(x - 2) = 0

Therefore, x = -3 or x = 2.

The limits of integration will be the x-values -3 and 2 because these are the points of intersection.

To evaluate the integral, we can use the formula for the area between two curves:
Area = ∫[a, b][f(x) - g(x)] dx

Therefore, the integral will be:
Area = ∫[-3, 2][(x^2 + 6x + 9) - (5(x+3))] dx
= ∫[-3, 2][(x^2 + 6x + 9) - (5x + 15)] dx
= ∫[-3, 2][x^2 + x - 6] dx

Now we can integrate:
= (1/3)x^3 + (1/2)x^2 - 6x | [-3, 2]
= [(1/3)(2)^3 + (1/2)(2)^2 - 6(2)] - [(1/3)(-3)^3 + (1/2)(-3)^2 - 6(-3)]
= [8/3 + 2 - 12] - [-27/3 + 9/2 + 18]
= [8/3 - 2/1 - 36/3] - [-9 + 9/2 + 18]
= [8/3 -6/3 - 36/3] - [-18 + 9/2 + 18]
= (-14/3) - (19/2)
= (-28/6) - (57/6)
= (-85/6)

Therefore, the area of the region bounded by f(x) = x^2 + 6x + 9 and g(x) = 5(x + 3) is -85/6.

2. To find the area of the region bounded by x = y^2 + 6, x = 0, y = -6, and y = 7, we need to set up the integral for each part of the region.

First, let's find the limits of integration:
The curve x = y^2 + 6 intersects with the y-axis when x = 0:
0 = y^2 + 6
y^2 = -6 (no real solutions)
Therefore, the lower limit of integration for y is -6.

The curve x = y^2 + 6 intersects with x = 0 when y = ±√(-6).
Since we are looking for the area between x = y^2 + 6 and x = 0, we only consider the positive square root:
y = √(-6)
Therefore, the upper limit of integration for y is √(-6).

The integral for the area is given by:
Area = ∫[-6, √(-6)](x - 0)dy = ∫[-6, √(-6)](y^2 + 6)dy

Now we can integrate:
= (1/3)y^3 + 6y | [-6, √(-6)]
= [(1/3)(√(-6))^3 + 6(√(-6))] - [(1/3)(-6)^3 + 6(-6)]
= [2√(-6) + 6√(-6)] - [-216/3 - 36]
= 8√(-6) + 72 - (-72)
= 8√(-6) + 144

Therefore, the area of the region bounded by x = y^2 + 6, x = 0, y = -6, and y = 7 is 8√(-6) + 144.

3. To find the volume of the solid generated by revolving the region bounded by the graphs of the equations about y = 3, we will use the method of cylindrical shells.

The integral for the volume is given by:
Volume = ∫[a, b] 2π(x)(y - 3) dx

We need to find the limits of integration by finding the points of intersection between the two curves.

The curves are x = y^2 + 6 and y = 3.

Setting them equal to each other:
y = 3
x = 3^2 + 6
x = 9 + 6
x = 15

Therefore, the limits of integration will be from y = 0 to y = 3.

The integral for the volume will be:
Volume = ∫[0, 3] 2π[(y^2 + 6)](y - 3) dy

Now we can integrate:
= 2π[(1/3)y^4 + 6y^2 - 3y^3]| [0, 3]
= 2π[(1/3)(3^4) + 6(3^2) - 3(3^3)] - 2π[(1/3)(0^4) + 6(0^2) - 3(0^3)]
= 2π[(1/3)(81) + 6(9) - 3(27)] - 0
= 2π[(27 + 54 - 81)] - 0
= 2π(0) - 0
= 0

Therefore, the volume of the solid generated by revolving the region bounded by the graphs of the equations about y = 3 is 0.

1. To find the area of the region bounded by the two curves, we need to find the points where the two curves intersect. Setting the two equations equal to each other, we have:

x^2 + 6x + 9 = 5(x + 3)

Simplifying the equation, we get:
x^2 + 6x + 9 = 5x + 15
x^2 + x - 6 = 0

Solving this quadratic equation, we find that x = -3 and x = 2 are the points of intersection. These will be the limits of integration.

To evaluate the integral, we need to integrate the difference of the two functions over the interval [-3, 2]. Since the upper curve is g(x) = 5(x + 3) and the lower curve is f(x) = x^2 + 6x + 9, the integral to find the area is:

∫ [g(x) - f(x)] dx

Substituting in the respective functions, we have:

∫ [5(x + 3) - (x^2 + 6x + 9)] dx

Now, we can evaluate this integral by distributing and simplifying:

∫ [5x + 15 - x^2 - 6x - 9] dx
∫ [-x^2 - x + 6] dx

Integrating term by term, we get:

(-1/3)x^3 - (1/2)x^2 + 6x + C

Evaluating this integral from -3 to 2, we substitute the limits of integration into the expression:

[(-1/3) * 2^3 - (1/2) * 2^2 + 6 * 2] - [(-1/3) * (-3)^3 - (1/2) * (-3)^2 + 6 * (-3)]

Simplifying further, we obtain:

[(8/3) - 2 + 12] - [(-27/3) - (9/2) - 18]
[8/3 - 2 + 12] - [-27/3 - 9/2 - 18]
[8/3 - 2 + 12] - [-9 - 9/2 - 18]
[8/3 - 2 + 12] - [-9 - 27/2]

Calculating these expressions, we get:

[50/3] - [-27 - 27/2]
[50/3] - [-54/2 - 27/2]
[50/3] - [-81/2]

To subtract fractions, we need a common denominator:

[50/3] - [-162/6]
[50/3] - [-162/6]

Now, we can subtract:

[50/3] + [162/6]
[100/6] + [162/6]
[262/6]

This can be simplified to:

131/3

Therefore, the area of the region bounded by f(x) = x^2 + 6x + 9 and g(x) = 5(x + 3) is 131/3 square units.

2. To find the area of the region bounded by the given curves, we can graph the equations and find the points of intersection. From the given equations, we have:
x = y^2 + 6, x = 0, y = -6, and y = 7

Graphing these equations, we can find that the curves intersect at y = 3 and y = -3. These will be the limits of integration.

Next, we need to integrate the difference of the two functions over the interval [-3, 3]. Since the upper curve is x = y^2 + 6 and the lower curve is x = 0, the integral to find the area is:

∫ [y^2 + 6 - 0] dy

Simplifying, we get:

∫ [y^2 + 6] dy

Integrating term by term, we have:

[y^3/3 + 6y]

Evaluating this integral from -3 to 3, we substitute the limits of integration into the expression:

[(3^3/3 + 6 * 3)] - [(-3^3/3 + 6 * -3)]
[(27/3 + 18)] - [(-27/3 - 18)]

Simplifying further, we obtain:

[(27/3 + 18)] - [(-27/3 - 18)]
[9 + 18] - [-9 - 18]
[9 + 18] - [-27]

The final result is:

27 - (-27) = 27 + 27 = 54

Therefore, the area of the region bounded by x = y^2 + 6, x = 0, y = -6, and y = 7 is 54 square units.

3. To find the volume of the solid generated by revolving the region bounded by the given graphs about y = 3, we can use the method of cylindrical shells.

First, we need to find the points of intersection of the two graphs. Since the equations of the given graphs are not provided, we need them to determine the points of intersection.

Once we have the equations, let's say x = f(y) and x = g(y), we can equate them to find the intersection points.

After finding the intersection points, we need to set up the integral that represents the volume of the solid generated by rotating the region between the two curves about y = 3.

The integral to find the volume is given by:

V = ∫ [2π(y - 3) * (g(y) - f(y))] dy

Here, (g(y) - f(y)) represents the height of a cylindrical shell at a given value of y, and (y - 3) represents the distance of the cylindrical shell from the axis of rotation.

Once the integral is set up, we can evaluate it and obtain the exact answer for the volume.