1. Use cylindrical shells to compute the volume of the solid formed by revolving the region bounded by y= 2x and y = x^2 -3 about x= 6.
2. Find the volume of the solid formed by revolving the region bounded by
y= e^x , y= e^-x, and y= 3 about y= 5.
1. To compute the volume of the solid formed by revolving the region bounded by y= 2x and y= x^2 - 3 about x= 6, we can use the method of cylindrical shells.
First, let's draw a rough sketch of the region and its axis of rotation to better understand the problem.
The region bounded by the curves y= 2x and y= x^2 - 3 is an area between two curves in the x-y plane.
Next, we need to find the range of x-values over which the curves intersect. By setting the two equations equal to each other, we have:
2x = x^2 - 3.
Rewriting this equation, we get:
x^2 - 2x - 3 = 0.
Factoring the quadratic equation, we have:
(x - 3)(x + 1) = 0.
This gives us two possible values for x: x = 3 and x = -1.
Now, we need to determine the height of the cylindrical shells at each value of x. The height of each shell is the difference in the y-values of the two curves at the corresponding x-value.
For x between -1 and 3, the height of the cylindrical shell is given by:
h(x) = y(2x) - y(x^2 - 3).
Substituting the equations for the curves, we have:
h(x) = 2x - (x^2 - 3) = -x^2 + 2x + 3.
Next, we need to find the radius of each cylindrical shell. The radius is the distance from the axis of rotation (x = 6) to the corresponding x-value.
r(x) = 6 - x.
Finally, we can write the volume of each cylindrical shell as:
V(x) = 2πr(x)h(x).
Integrating this expression over the range of x-values (-1 to 3), we can find the total volume of the solid:
V = ∫(from -1 to 3) 2π(6-x)(-x^2 + 2x + 3) dx.
Computing this integral will give us the desired volume of the solid formed by revolving the given region about x= 6.
2. To find the volume of the solid formed by revolving the region bounded by y= e^x, y= e^-x, and y= 3 about y= 5, we can again use the method of cylindrical shells.
First, let's draw a rough sketch of the region and its axis of rotation to better understand the problem.
The region bounded by the curves y= e^x, y= e^-x, and y= 3 is an area between three curves in the x-y plane.
Next, we need to find the range of x-values over which the curves intersect. By setting the equations equal to each other, we have:
e^x = e^-x and e^x = 3.
Solving these equations, we find x = 0 for the first equation and x = ln(3) for the second equation.
Now, we need to determine the height of the cylindrical shells at each value of x. The height of each shell is the difference between the upper and lower curves at the corresponding x-value.
For x between 0 and ln(3), the height of the cylindrical shell is given by:
h(x) = 3 - e^x - e^-x.
Next, we need to find the radius of each cylindrical shell. The radius is the distance from the axis of rotation (y = 5) to the corresponding y-value.
r(x) = 5 - y(x).
Finally, we can write the volume of each cylindrical shell as:
V(x) = 2πr(x)h(x).
Integrating this expression over the range of x-values (0 to ln(3)), we can find the total volume of the solid:
V = ∫(from 0 to ln(3)) 2π(5 - y(x))(3 - e^x - e^-x) dx.
Computing this integral will give us the desired volume of the solid formed by revolving the given region about y= 5.