3). The shaded region is bounded by the y-axis and the graphs of y=1+√x, y=2. Find the volume of the solid obtained by rotating this region around the x-axis.
Answer choices: 7/6pi, 4/3pi, 11/6pi, 5/3pi, 13/6pi, 5/6pi
4). Find the area of the region bounded by y=x^2-6x+7, y=7+6x-x^2
3. I don't see your graph but sketched it.
V= π∫(4 - (1+√x)^2 dx from x = 0 to 1
= π∫ (4 - 1 - 2√x - x) dx from 0 to 1
= π [ 3x - (4/3)x^(3/2) - (1/2)x^2 ] from 0 to 1
= π [ 3 - 4/3 - 1/2 - ( 0 )]
= π[7/6]
= 7π/6 , which is the first choice
4. first find their intersection:
x^2 - 6x + 7 = 7 + 6x - x^2
2x^2 - 12x = 0
x^2 - 6x = 0
x(x-6) = 0
so x = 0 or x = 6
area = ∫7 + 6x - x^2 - (x^2 - 6x + 7) dx from 0 to 6
= ∫( -2x^2 + 12x) dx from 0 to 6
= [(-2/3)x^3 + 6x^2] from 0 to 6
= ((-2/3)(216) + 216 - 0)
= 72
Ah, I see you're in for a mathematical challenge! Let's put on our thinking caps and solve these problems with a touch of humor, shall we?
3) To find the volume of the solid obtained by rotating the shaded region around the x-axis, imagine you're at a circus fair spinning around on a carousel. Wheee! Now, let's do some math. We can use the disk method to calculate this volume.
First, we need to find the points of intersection between the two curves y = 1 + √x and y = 2. To do this, we set the two equations equal to each other: 1 + √x = 2. Solving for x, we find x = 1.
Now that we have our boundaries, we integrate to get the volume. The formula for the volume using the disk method is V = π * ∫(lower-boundary)(upper-boundary) [f(x)]^2 dx.
Integrating from x = 0 to x = 1, we have V = π * ∫(0 to 1) [1 + √x]^2 dx.
Simplify the equation and integrate: V = π * ∫(0 to 1) [1 + 2√x + x] dx.
Now, integrate each term separately: V = π * ∫(0 to 1) (1 + 2√x + x) dx.
Integrating, we get V = π * [x + 4/3 * x^(3/2) + 1/2 * x^2] evaluated from 0 to 1.
Plugging in the limits, we get V = π * [1 + 4/3 + 1/2 - 0 - 0 - 0] = π * [11/6] = 11/6π.
So, the volume of the solid obtained by rotating the shaded region around the x-axis is 11/6π. Ta-da!
4) Ah, another question on finding the area of a region bounded by curves. It's time for some geometric calculations infused with a dash of comedy!
To find the area bounded by the curves y = x^2 - 6x + 7 and y = 7 + 6x - x^2, we need to find their points of intersection first. Let's call these points A and B for simplicity.
Setting the two equations equal to each other, we get: x^2 - 6x + 7 = 7 + 6x - x^2.
Combining like terms and simplifying, we have: 2x^2 - 12x = 0.
Factoring out 2x, we get: 2x(x - 6) = 0.
Solving for x, we have x = 0 (point A) and x = 6 (point B).
To find the area, we integrate the function f(x) = (y2 - y1) dx, where y1 = x^2 - 6x + 7 and y2 = 7 + 6x - x^2, with the limits of integration from x = 0 to x = 6.
Integrating, we get the area: A = ∫(0 to 6) [(7 + 6x - x^2) - (x^2 - 6x + 7)] dx.
Simplifying, we have: A = ∫(0 to 6) (14x - 2x^2) dx.
Now, let's integrate each term separately: A = [7x^2 - 2/3 * x^3] evaluated from 0 to 6.
Plugging in the limits, we get: A = [7(6)^2 - 2/3 * (6)^3] - [7(0)^2 - 2/3 * (0)^3].
Simplifying further, we have: A = 252 - 0 = 252 square units.
Therefore, the area of the region bounded by y = x^2 - 6x + 7 and y = 7 + 6x - x^2 is 252. Voila!
To find the volume of the solid obtained by rotating the shaded region around the x-axis, we can use the method of cylindrical shells.
To start, let's first find the points of intersection between the two curves: y = 1 + √x and y = 2.
Setting the two equations equal to each other, we have:
1 + √x = 2
Solving for x, we get:
√x = 1
x = 1
So the curves intersect at the point (1, 2).
Next, let's find the limits of integration for the volume integral. The shaded region is bounded by the y-axis, which means the limits of integration for x are from 0 to 1.
Now, let's consider a representative strip of the shaded region with width Δx. When this strip is rotated around the x-axis, it generates a cylindrical shell with radius x and height (2 - (1 + √x)).
The volume of each cylindrical shell is given by the formula V = 2πrhΔx, where r is the radius and h is the height.
Substituting the values, we have:
V = 2πx(2 - (1 + √x))Δx
To find the total volume, we need to integrate this expression from x = 0 to 1:
V = ∫[0,1] 2πx(2 - (1 + √x)) dx
Simplifying the expression inside the integral:
V = ∫[0,1] 2πx(1 - √x) dx
Next, we need to integrate this expression. Expanding the integrand:
V = ∫[0,1] 2πx - 2πx√x dx
Integrating term by term:
V = π[x^2 - (2/3)x^(3/2)] evaluated from 0 to 1
Substituting the limits of integration:
V = π[(1)^2 - (2/3)(1)^(3/2)] - π[(0)^2 - (2/3)(0)^(3/2)]
V = π[1 - (2/3)(1)] - π[0 - 0]
V = π[1 - 2/3]
V = π/3
Therefore, the volume of the solid obtained by rotating the shaded region around the x-axis is π/3.
Answer: The correct answer is π/3.
To find the volume of the solid obtained by rotating the shaded region around the x-axis, we can use the method of cylindrical shells.
1) Start by graphing the two given equations: y = 1 + √x and y = 2.
2) To find the bounds for integration, we need to find the x-values where these two curves intersect. Set the equations equal to each other:
1 + √x = 2
√x = 1
x = 1
So, the intersection point is x = 1. This will be our upper bound for integration.
3) Now, we need to find the bounds for the lower limit of integration. The shaded region is bounded by the y-axis, which corresponds to x = 0. Therefore, the lower limit of integration is x = 0.
4) We can write the equation for the shaded region as y = 2 - (1 + √x). Simplifying this equation, we get y = 1 - √x.
5) To set up the integral for finding the volume, we need to find the height of the cylindrical shell at each x-value and multiply it by the circumference of the shell (which is 2πx). The height of the cylindrical shell is the difference between the upper curve (y = 2) and the lower curve (y = 1 - √x).
Volume = ∫[0, 1] 2πx * (2 - (1 - √x)) dx
6) Simplify the integrand:
Volume = ∫[0, 1] (2πx - 2πx√x + 2π√x) dx
7) Integrate the expression with respect to x using the limits of integration 0 and 1.
After integrating and simplifying, you will get the volume of the solid.
Regarding the second question:
1) Start by graphing the given equation: y = x^2 - 6x + 7 and y = 7 + 6x - x^2.
2) To find the area of the region bounded by these curves, we need to find the x-values where they intersect. Set the two equations equal to each other:
x^2 - 6x + 7 = 7 + 6x - x^2
2x^2 - 12x = 0
2x(x - 6) = 0
So, the x-values are x = 0 and x = 6. These will be our bounds for integration.
3) We can write the equation for the shaded region as y = (7 + 6x - x^2) - (x^2 - 6x + 7). Simplifying this equation, we get y = 12x - x^2.
4) Set up the integral for finding the area, which is given by ∫[0, 6] (12x - x^2) dx.
5) Integrate the expression with respect to x using the limits of integration 0 and 6.
After integrating and simplifying, you will get the area of the region bounded by the two given curves.