1. Find the volume V of the solid obtained by rotating the region bounded by the given curves about the specified line.
y = ln(5x), y = 1, y = 3, x = 0; about the y-axis
2. Use the method of cylindrical shells to find the volume V generated by rotating the region bounded by the given curves about the specified axis.
y = 32 − x2, y = x2; about x = 4
3. Use the method of cylindrical shells to find the volume V generated by rotating the region bounded by the given curves about y = 8.
64y = x3, y = 0, x = 8
4. A tank is full of water. Find the work required to pump the water out of the spout. Use the fact that water weighs 62.5 lb/ft3. (Assume a = 6 ft, b = 8 ft, and c = 12 ft.)
5.If 2.4 J of work are needed to stretch a spring from 9 cm to 13 cm and 4 J are needed to stretch it from 13 cm to 17 cm, what is the natural length of the spring?
Oops. I'm sure you caught my error in #1.
v = π(e/5)^2(2 + ∫[e/5, (e/5)^3] 2πrh dx
1. To find the volume of the solid obtained by rotating the region bounded by the given curves about the y-axis, you can use the method of cylindrical shells.
First, you need to determine the limits of integration. In this case, the region is bounded by y = 1 and y = 3.
Next, set up the integral:
V = ∫(2πx)(y) dx
Since we are rotating about the y-axis, x is the variable of integration. We need to rewrite the equations in terms of x.
Using the equation y = ln(5x), solve for x:
e^y = 5x
x = e^y/5
So the integral becomes:
V = ∫(2π(e^y/5))(y) dy
Now, evaluate the integral using the given limits of integration, y = 1 and y = 3:
V = ∫[1,3](2π(e^y/5))(y) dy
Calculate the definite integral to find the volume V.
2. To find the volume V generated by rotating the region bounded by the given curves about the line x = 4, we can use the method of cylindrical shells.
First, determine the limits of integration. In this case, the region is bounded by the curves y = 32 - x^2 and y = x^2.
Next, set up the integral:
V = ∫(2πr)(h) dx
Since we are rotating about the line x = 4, the variable of integration is x.
To find r (the radius), subtract x from the line of rotation: r = 4 - x.
To find h (the height), subtract the upper curve y = 32 - x^2 from the lower curve y = x^2: h = (32 - x^2) - (x^2) = 32 - 2x^2.
Now, substitute r and h into the integral:
V = ∫(2π(4 - x))(32 - 2x^2) dx
Now, evaluate the integral using the given limits of integration, which depend on the points of intersection between the curves.
Calculate the definite integral to find the volume V.
3. To find the volume V generated by rotating the region bounded by the given curves about the line y = 8, we can use the method of cylindrical shells.
First, determine the limits of integration. In this case, the region is bounded by y = 0 and y = 64/x^3.
Next, set up the integral:
V = ∫(2πr)(h) dy
Since we are rotating about the line y = 8, the variable of integration is y.
To find r (the radius), subtract 8 from the y-coordinate: r = 8 - y.
To find h (the height), solve the equation 64y = x^3 for x: x = (64y)^(1/3).
Now, substitute r and h into the integral:
V = ∫(2π(8 - y))((64y)^(1/3)) dy
Now, evaluate the integral using the given limits of integration, which are y = 0 and y = 8 (since these are the limits of the region bounded by y = 0 and y = 64/x^3).
Calculate the definite integral to find the volume V.
4. To find the work required to pump the water out of the spout, you need to calculate the weight of the water.
The weight of water is given as 62.5 lb/ft^3.
The volume V of the tank is given by the formula V = a * b * c, where a, b, and c are the dimensions of the tank.
The weight of the water can be calculated by multiplying the volume by the density:
W = V * density
Substituting the given values, we have:
W = (6 ft * 8 ft * 12 ft) * (62.5 lb/ft^3)
Calculate the expression to find the work required to pump the water out of the spout.
5. To find the natural length of the spring, we can use the concept of work done.
The work done in stretching a spring is given by the equation:
Work = (1/2) * k * (x^2 - x0^2)
Where k is the spring constant, x is the final displacement, and x0 is the initial displacement.
In this case, we are given two sets of values for the work and displacements:
1. Work = 2.4 J, x = 13 cm, and x0 = 9 cm
2. Work = 4 J, x = 17 cm, and x0 = 13 cm
We can set up two equations using these values:
2.4 = (1/2) * k * (13^2 - 9^2)
4 = (1/2) * k * (17^2 - 13^2)
Simplify and solve these equations simultaneously to find the value of k (the spring constant).
Once you have the value of k, use one of the equations (e.g., the first one) to solve for the natural length x0.
Plug in the values of k, x, and the computed x0 to find the natural length of the spring.
another homework dump?
Looks like you need to show some work. I'll get you started.
#1. y = ln(5x), y = 1, y = 3, x = 0; about the y-axis
using shells of thickness dx,
v = 2(e/5) + ∫[e/5, (e/5)^3] 2πrh dx
where r = x and h=3-ln(5x)
using discs of thickness dy,
v = ∫[1,3] πr^2 dy
where r = x = e^(y/5)
#2.
v = ∫[-4,4] 2πrh dx
where r = 4-x and h = 32-x^2 - x^2
#3.
v = ∫[0,8] 2πrh dy
where r = 8-y and h = x = 4∛y
#4.
No idea what a,b,c are supposed to be. But, find where the center of mass is for the tank, and the work required is just weight * height_lifted
and the weight is the volume * density
#5.
work = 1/2 kx^2
where x is the distance stretched.
So, if the natural length is a, we have
k/2 ((13-a)^2 - (9-a)^2) = 2.4
k/2 ((17-a)^2 - (13-a)^2) = 4
Now divide and solve for a.