1. Let R be the region in the first quadrant enclosed by the graphs of y=4-X , y=3x , and the y-axis.
a. Find the area of region R.
b. Find the volume of the solid formed by revolving the region R about the x-axis.
a. Finding the area of region R is as easy as pie! Well, not exactly, but let's give it a try!
To find the area of region R, we need to find the points where the graphs intersect. By setting the equations equal to each other, we can solve for x:
4 - x = 3x
Now, add x to both sides:
4 = 4x
Divide by 4:
x = 1
Now that we know the intersection point, we can find the area of region R by integrating the difference between the two functions from x = 0 to x = 1:
∫ (4 - x) - (3x) dx from 0 to 1
Integrating this will give us the area of region R!
b. Now it's time to take region R for a spin! We can find the volume of the solid formed by revolving region R about the x-axis using the magical method of cylindrical shells.
The volume V can be calculated using the formula:
V = 2π∫ (x)(y) dx from 0 to 1
By substituting the values of x and y into the formula, we can evaluate the integral and find the volume.
So, get your math cap on and calculate away! Just remember to have some popcorn ready because integrals can be quite entertaining!
To find the area of region R, we need to find the points where the graphs y=4-X, y=3x and the y-axis intersect each other.
a. Finding the points of intersection:
1. Set y=4-X equal to y=3x:
4-X = 3x
2. Rearrange the equation to solve for x:
X + 3x = 4
4x = 4
x = 1
3. Substitute the value of x into either equation to find y:
y = 3(1)
y = 3
So the point of intersection is (1, 3).
4. To find the other point of intersection, set y=3x equal to y=0 (the y-axis):
3x = 0
5. Solve for x:
x = 0
The other point of intersection is (0, 0).
b. The graphs y=4-X and y=3x intersect at two points: (1, 3) and (0, 0).
To find the area of region R, we need to find the area between the curves y=4-X and y=3x from x=0 to x=1. We can do this by integrating the difference between the two curves.
1. Set up the integral for the area:
A = ∫[0-1] (4-X - 3x) dx
2. Integrate the function:
A = ∫[0-1] (4 - 3x - X) dx
A = ∫[0-1] (4 - 4x) dx
3. Integrate and evaluate the integral:
A = [4x - 2x^2] [0-1]
A = (4(1) - 2(1)^2) - (4(0) - 2(0)^2)
A = 4 - 2 - 0 + 0
A = 2
Therefore, the area of region R is 2 square units.
To find the volume of the solid formed by revolving region R about the x-axis, we need to use the method of cylindrical shells.
1. Set up the integral for the volume:
V = 2π ∫[0-1] (x)(4-x) dx
2. Integrate the function:
V = 2π ∫[0-1] (4x - x^2) dx
3. Integrate and evaluate the integral:
V = 2π [(2x^2 - (1/3)x^3)] [0-1]
V = 2π [(2(1)^2 - (1/3)(1)^3) - (2(0)^2 - (1/3)(0)^3)]
V = 2π [(2 - (1/3)) - (0 - 0)]
V = 2π [(6/3 - 1/3)]
V = 2π (5/3)
V = (10/3)π
Therefore, the volume of the solid formed by revolving region R about the x-axis is (10/3)π cubic units.
To find the area of region R, we need to determine the points of intersection between the three given graphs.
To begin, let's find the points of intersection between y = 4 - x and y = 3x:
Set the two equations equal to each other:
4 - x = 3x
Add x to both sides:
4 = 4x
Divide both sides by 4:
x = 1
Substitute x = 1 into either equation to find the corresponding y-value:
y = 4 - 1
y = 3
So, the point of intersection between y = 4 - x and y = 3x is (1, 3).
Next, let's find the points of intersection between y = 3x and the y-axis:
Set x = 0 in y = 3x:
y = 3 * 0
y = 0
So, the point of intersection between y = 3x and the y-axis is (0, 0).
Now that we have identified the points of intersection, we can divide region R into two separate shapes: a triangle (formed by y = 3x, y = 4 - x, and the y-axis) and a trapezoid (formed by y = 3x, y = 4 - x, and the x-axis).
To calculate the area of the triangle, we use the formula:
Area = (base * height) / 2
In this case, the base is the length between x = 0 and x = 1 (which is 1 - 0 = 1), and the height is the y-coordinate of the point of intersection between y = 4 - x and y = 3x (which is 3). Thus, the area of the triangle is:
Area of triangle = (1 * 3) / 2
= 3/2
= 1.5
To calculate the area of the trapezoid, we use the formula:
Area = (base1 + base2) * height / 2
In this case, the bases are the lengths between x = 0 and x = 1 (which is 1 - 0 = 1) and the x-coordinate of the point of intersection between y = 4 - x and y = 3x (which is 1 - 0 = 1). The height is the y-coordinate of the point of intersection between y = 4 - x and y = 3x (which is 3). Thus, the area of the trapezoid is:
Area of trapezoid = (1 + 1) * 3 / 2
= 2 * 3 / 2
= 6 / 2
= 3
Now, to find the total area of region R, we add the areas of the triangle and trapezoid:
Total Area of region R = Area of triangle + Area of trapezoid
= 1.5 + 3
= 4.5
Therefore, the area of region R is 4.5 square units.
To find the volume of the solid formed by revolving region R about the x-axis, we will use the method of cylindrical shells.
For each x-value between 0 and 1, the height of the cylinder is given by the formula y = 3x - (4 - x) = 4x - 4.
The radius of the cylinder at each x-value is the value of x itself.
The volume of each cylinder is given by the formula V = 2πrh, where r is the radius and h is the height.
Therefore, the volume of the solid formed by revolving region R about the x-axis is given by the integral:
V = ∫[0,1] 2π(4x - 4)x dx
Simplifying the integral, we have:
V = 2π ∫[0,1] (4x^2 - 4x) dx
V = 2π ( (4/3)x^3 - 2x^2 ) |[0,1]
V = 2π ( (4/3)(1)^3 - 2(1)^2 ) - 2π ( (4/3)(0)^3 - 2(0)^2 )
V = 2π ( 4/3 - 2 )
V = 2π ( 4/3 - 6/3 )
V = 2π ( -2/3 )
V = -4π/3
Therefore, the volume of the solid formed by revolving region R about the x-axis is -4π/3 cubic units.
need the intersection ....
3x = 4-x
4x=4
x = 1
height of revolved region = radius of rotation
= 4-x - 3x = 4 - 4x
area = ∫(4-4x) dx from 0 to 1
= [4x - 2x^2] from 0 to 1
= (4 - 2) - 0
= 2
volume = π∫ (4-4x) dx from 0 to 1
= π∫(16 - 32x + 16x^2) dx from 0 to 1
= π[16x - 16x^2 + (16/3)x^3 ] from 0 to 1
= π( 16 - 16 + 16/3 - 0)
= 16π/3
check my arithmetic