(1) The region in the x-y plane bounded by the curve y=x^2, the line x=0 and the line x=5 and the x-axis. find the volume generated
(2) Assume that a force of (3x^2-x+1)N is pulling an object at each point of the x-axis. find the work done in moving the object from x=1 to x=4
1) THE differential area dA will be y dx or x^2 dx, integrated from x=0 to 5. Now integrate that around the x axis, by rotation. Area=INT Area dTheta where theta runs from 0 to 2PI
2) Work= INT force dx
= INT(3x^2-x+1)dx from x=1 to 4
To find the volume generated in question (1), we can use the method of cross-sections.
Step 1: Sketch the region bounded by the curve y=x^2, the line x=0 and the line x=5, and the x-axis. This region looks like a parabolic shape with a base between x=0 and x=5.
Step 2: Divide the region into infinitesimally small rectangles or squares along the x-axis. Each rectangle or square will have a width of dx.
Step 3: Determine the height of each rectangle or square by evaluating the corresponding y-value on the curve y=x^2. The height of each rectangle is given by y=x^2.
Step 4: Calculate the area of each rectangle or square by multiplying the width (dx) by the height (y=x^2). The area of each rectangle is given by A = x^2 * dx.
Step 5: Integrate the areas of all the rectangles or squares from x=0 to x=5 to find the total volume. The volume is given by V = ∫[0,5] (x^2 * dx).
To calculate the integral, apply the power rule for integration: ∫x^n dx = (1/(n+1)) * x^(n+1).
Using this rule, the volume can be calculated as follows:
V = ∫[0,5] (x^2 * dx)
= [(1/3) * x^3] | from 0 to 5
= (1/3) * (5^3 - 0^3)
= (1/3) * 125
= 41.67 cubic units
Therefore, the volume generated by the region is approximately equal to 41.67 cubic units.
To find the work done in question (2), we can use the method of integration.
Step 1: Identify the force function given as f(x) = 3x^2 - x + 1 N.
Step 2: Determine the displacement of the object by finding the difference between the final position and the initial position. In this case, the final position is x=4 and the initial position is x=1, so the displacement is Δx = 4 - 1 = 3.
Step 3: Integrate the force function with respect to x over the range from x=1 to x=4 to calculate the work done. The work done is given by W = ∫[1,4] (3x^2 - x + 1) dx.
Using the power rule for integration, the integral can be evaluated as follows:
W = ∫[1,4] (3x^2 - x + 1) dx
= [x^3 - (1/2)x^2 + x] | from 1 to 4
= ((4^3 - (1/2)*4^2 + 4) - (1^3 - (1/2)*1^2 + 1))
= (64 - 8 + 4) - (1 - 0.5 + 1)
= 60 - 0.5
= 59.5 N·m
Therefore, the work done in moving the object from x=1 to x=4 is equal to 59.5 N·m.