Question

Use the method of cylindrical shells to find the volume V of the solid obtained by rotating the region bounded by the given curves about the x-axis. x + y = 7, x = 16 − (y − 3)2

Answer 1

To find the volume using cylindrical shells, we set up an integral based on the region between the two given curves, after expressing one in terms of the other. We then integrate the volume of each cylindrical shell across the bounded region to find the total volume.

Step-by-step explanation:

Finding Volume using Cylindrical Shells

When finding the volume of a solid of revolution using the method of cylindrical shells, we consider infinitesimally thin cylindrical shells at a distance r from the axis of rotation (in this case, the x-axis) with height given by the function representing the boundary of the region. The general formula for the volume of a thin shell of radius r, height h, and thickness dr is: V = 2πrh dr. Integrating this expression from the lower to the upper bounds of r gives us the total volume.

For the curves x + y = 7 and x = 16 − (y − 3)^2, we first need to find the intersection points and express y as a function of x or vice versa for the shell method. Then, we'll set up an integral using the height of the cylindrical shell as the difference between these two functions. Lastly, we'll integrate this with respect to the radius across the interval defined by the region bounded by these curves.

Keep in mind that it's essential to set up the integral correctly based on the curves and the axis of rotation. For x-axis rotation, we typically integrate with respect to y, and our radius is a function of y as well.

The provided information such as the volume being half that of the cube (4r³) or involving a multiplier of 7 does not directly apply to this problem unless determined by the specific bounds of integration after setting up the problem properly with the given functions.

Answer 2

`v= (2401\pi)/(6)`

Step-by-step explanation:

I put the equations in a plotter, you can see the rotation region below. Now, as we are rotationg about the x-axis we are going to use the equations and the limit points in terms of y. Now, the volume is

`v = 2\pi\int\limits^a_b {p(y)h(y)} \, dy` where p(y) is the distance from the rotation axis and the differential, a and b are the limit points of the region and h(y) is the height of the region. In this case p(y)= y, a=0, b=7 and h(y) = 16-(y-3)^2-(7-y) (right minus left). So,

`v = 2\pi\int\limits^0_7 {y(16-(y-3)^(2)-7+y)} \, dy`

`v = 2\pi\int\limits^0_7 {y(9-(y-3)^(2)+y)} \, dy`

`v = 2\pi\int\limits^0_7 {y(9-(y^(2)-6y+9)+y)} \, dy`

`v = 2\pi\int\limits^0_7 {y(9-y^(2)+6y-9+y)} \, dy`

`v = 2\pi\int\limits^0_7 {y(-y^(2)+7y)} \, dy`

`v = 2\pi\int\limits^0_7 {-y^(3)+7y^(2)} \, dy`

`v= 2\pi (-(y^(4))/(4)+(7y^(3))/(3))^(7)_0`

`v= 2\pi (-(7^(4))/(4)+(7*7^(3))/(3))`

`v= 2\pi (-(2401)/(4)+(2401)/(3))`

`v= 2\pi ((2401)/(12))`

`v= (2401\pi)/(6)`.