Final answer:
The volume of the solid generated by revolving the region about the x-axis is 32π/5, and the volume of the solid generated by revolving the region about the y-axis is π/2.
Step-by-step explanation:
To find the volume of the solid generated by revolving the region bounded by y = x^2, y = 0, and x = 2 about the x-axis, we can use the method of disks or washers. Since we are revolving around the x-axis, the cross-sections will be circular disks. The radius of each disk is the value of y = x^2, so the volume of each disk is V = π(x^2)^2 = πx^4. To find the total volume, we integrate from x = 0 to x = 2:
- Calculate the volume of one disk: V = πx^4
- Integrate the volume function over the interval [0, 2]: ∫(πx^4)dx
- Evaluate the definite integral: V = ∫πx^4dx = π/5 * x^5]
By substituting the limits of integration, we get:
V = π/5 * (2^5 - 0^5) = π/5 * 32 = 32π/5.
Therefore, the volume of the solid generated by revolving the region about the x-axis is 32π/5.
To find the volume of the solid generated by revolving the same region about the y-axis, we need to use the method of cylindrical shells. Since we are revolving around the y-axis, the cross-sections will be cylindrical shells. The radius of each shell is x, and the height is y = x^2. The volume of each shell is V = 2πxy dx = 2πx * x^2 dx = 2πx^3 dx. To find the total volume, we integrate from x = 0 to x = 2:
- Calculate the volume of one shell: V = 2πx^3
- Integrate the volume function over the interval [0, 2]: ∫(2πx^3)dx
- Evaluate the definite integral: V = ∫2πx^3dx = π/2
Therefore, the volume of the solid generated by revolving the region about the y-axis is π/2.