Sketch the region enclosed by the lines x=0 x=6 y=2 and y=6. Identify the vertices of the region. Revolve the region around the y-axis. Identify the solid formed by the revolution calculate the volume of the solid. Leave the answer in terms of pi.
To sketch the region enclosed by the lines x=0, x=6, y=2, and y=6, we can plot these lines on a Cartesian plane:
```
Y-axis
|
| (6,6)
|______|______ x-axis
| (6,2)
|
|
|
| (0,2)
|______|______
(0,0)
```
The region enclosed by these lines is a rectangle with its lower left vertex at (0,2) and its upper right vertex at (6,6).
To revolve this region around the y-axis, we obtain a solid called a cylindrical shell.
The height of the shell is the difference in y-coordinates of the vertices, which is 6 - 2 = 4.
The radius of the shell is the x-coordinate at the outer edge, which is 6.
To calculate the volume of the solid, we can use the formula for the volume of a cylindrical shell:
Volume = 2π * radius * height
Volume = 2π * (6) * (4)
Volume = 48π
Hence, the volume of the solid formed by revolving the region around the y-axis is 48π.
To sketch the region enclosed by the given lines, we can plot them on a coordinate plane.
Since x=0 and x=6 are vertical lines parallel to the y-axis, they create the left and right boundaries of the region. Similarly, since y=2 and y=6 are horizontal lines parallel to the x-axis, they form the lower and upper boundaries of the region.
Now, let's identify the vertices of the region:
- The left boundary intersects with the lower boundary at the point (0, 2).
- The left boundary intersects with the upper boundary at the point (0, 6).
- The right boundary intersects with the lower boundary at the point (6, 2).
- The right boundary intersects with the upper boundary at the point (6, 6).
Next, to revolve the region around the y-axis, we will create a solid called a cylindrical shell. This solid is formed by rotating the region enclosed by the lines around the y-axis.
To calculate the volume of the solid obtained, we can use the method of cylindrical shells:
V = 2π ∫[a, b] (x * h(x)) dx
In this case, a = 2 and b = 6, since those are the y-values of the lower and upper boundaries, which determine the height of the region.
Now, we need to express x as a function of y to obtain the height of each cylindrical shell. Since we are revolving the region around the y-axis, x remains constant for each given y-value. Therefore, x = y for all points within the region.
Substituting these values into the volume formula:
V = 2π ∫[2, 6] (y * (6 - 2)) dy
V = 2π ∫[2, 6] (4y) dy
Solving the integral:
V = 2π * [2y^2] [2, 6]
V = 2π * (2 * 6^2 - 2 * 2^2)
V = 2π * (72 - 8)
V = 2π * 64
V = 128π
Therefore, the volume of the solid formed by revolving the region around the y-axis is 128π cubic units.
you can easily see that the region is a rectangle. Rotate a rectangle around one of its sides and you get a cylinder. Then it's easy.
See my work at
http://www.jiskha.com/display.cgi?id=1462581546