Washer Method:
y=x^3, y=1 ,x=2 aroud the Y axis
To find the volume of the solid generated by rotating the region bounded by the curve y = x^3, the line y = 1, and the line x = 2 around the y-axis, we can use the washer method.
The washer method involves integrating the difference between the outer and inner radii of infinitesimally small cylindrical washers that make up the solid.
To begin, let's sketch the graph of the region bounded by the given curves:
1. Sketch the curve y = x^3.
2. Draw the horizontal line y = 1.
3. Draw the vertical line x = 2.
The region bounded by these curves will have a shape similar to a right triangular prism.
To find the outer radius (R) and inner radius (r) of the washers, we need to consider the distance between the y-axis and the curves at each y-coordinate within the bounded region.
1. For the outer radius (R):
- At any y-coordinate, the distance between the y-axis and the curve y = x^3 will be x. Hence, R = x.
2. For the inner radius (r):
- At the y-coordinate where y = 1 intersects with y = x^3, we can solve for x by setting x^3 = 1. From this, we find that x = 1.
- As we move from y = 1 down to y = 0, the distance between the y-axis and the curve y = x^3 decreases. At any y-coordinate within the bounded region, this distance will be x.
- Therefore, r = x.
We can now set up the integral to find the volume of the solid using the washer method:
∫[a, b] π(R^2 - r^2) dy
In our case, a = 0 (the lower y-coordinate of the region) and b = 1 (the upper y-coordinate of the region).
∫[0, 1] π(x^2 - x^2) dy
Simplifying the integral:
∫[0, 1] π(0) dy
Since the difference between the outer and inner radii is 0, the integral evaluates to 0.
Therefore, the volume of the solid generated by rotating the region around the y-axis is 0 cubic units.