Consider the region bounded by the parabola y=x^2 and the line y=16 .
(a) What is the volume of the solid generated when revolving this region about the line y=16 ?
first sketch it. then determine where the y=16 line intersects (-4,16)(4,16)
Now, lets integrate from the parabola to the line
dArea=dx*(16-y)=(16-x^2)dx
area= INT (16-x^2)dx from x=-4 to 4
area= [16x-1/3 x^3] over the limits
area= 16*4-1/3 4^3 - 16(-4)+1/3 (-4)^3
=2(16*4 - 1/3 4^3)
and you can finish it.
You could have recognized symettry, and just integrated from x=0 to 4, then doubled the area.
To find the volume of the solid generated when revolving the region bounded by the parabola y = x^2 and the line y = 16 about the line y = 16, we can use the method of cylindrical shells.
The cylindrical shell method involves taking thin cylindrical shells perpendicular to the axis of revolution and summing up their volumes to find the total volume of the solid.
To begin, let's first find the points of intersection between the parabola y = x^2 and the line y = 16. Setting the two equations equal, we have:
x^2 = 16
Taking the square root of both sides, we get:
x = ±√16
x = ±4
So the points of intersection are (4, 16) and (-4, 16).
Next, we need to express the equations in terms of x, as we will be integrating with respect to x. The equation of the parabola y = x^2 can be rewritten as x = √y.
Now, let's consider a small vertical strip of width dx and height y (between x and x + dx) on the curve of the parabola. If we revolve this strip around the line y = 16, it will generate a cylindrical shell of infinitesimal thickness.
The volume of this cylindrical shell can be calculated as the product of its height (which is the circumference of the shell) and its width (dx).
The height of the cylindrical shell is given by 2πy, as it represents the circumference of the shell at each point x.
So the volume of the cylindrical shell is given by:
dV = 2πy * dx
Now, we need to find the limits of integration for x. The parabola intersects the line y = 16 at x = ±4, so these will be our limits.
To find the total volume, we integrate the equation for the cylindrical shell volume over the interval from x = -4 to x = 4:
V = ∫[from -4 to 4] 2πy * dx
Substituting x = √y, we can rewrite the equation as:
V = ∫[from 16 to 256] 2πy * d(√y)
Now, we can simplify the integral:
V = 2π ∫[from 16 to 256] y^(1/2) dy
Using the power rule of integration, we integrate:
V = 2π * [(2/3) * y^(3/2)] |[from 16 to 256]
Putting in the limits of integration:
V = 2π * [(2/3) * (256^(3/2) - 16^(3/2))]
Evaluating the expression gives us the volume of the solid generated when revolving this region about the line y = 16.