Using washers, find the volume of the solid formed by rotating y = x^2 from x = 0 to 2 about the x-axis.

Using washers, find the volume of the solid formed by rotating y = x^2 from x = 0 to 2 about the x-axis.

• A. 32π/5
• B. 64π/5
• C. 16π/5
• D. 8π/5

Answer: (B)

Rotating the region under the curve y = x^2 from x = 0 to 2 about the x-axis creates disks whose radius at each x is the distance from the x-axis to the curve, which is y = x^2. The inner radius is 0 since the region touches the axis. So the volume is the washer (disk) method integral: V = π ∫ from 0 to 2 of [radius]^2 dx = π ∫_0^2 (x^2)^2 dx = π ∫_0^2 x^4 dx. Compute the integral: ∫_0^2 x^4 dx = [x^5/5]_0^2 = 32/5. Therefore V = π · 32/5 = 32π/5. So the volume is 32π/5. If a different value appears in a key, it might come from a different interpretation of the region, but with the stated setup, 32π/5 is the correct result.

Rotating the region under the curve y = x^2 from x = 0 to 2 about the x-axis creates disks whose radius at each x is the distance from the x-axis to the curve, which is y = x^2. The inner radius is 0 since the region touches the axis. So the volume is the washer (disk) method integral: V = π ∫ from 0 to 2 of [radius]^2 dx = π ∫_0^2 (x^2)^2 dx = π ∫_0^2 x^4 dx.

Compute the integral: ∫_0^2 x^4 dx = [x^5/5]_0^2 = 32/5. Therefore V = π · 32/5 = 32π/5.

So the volume is 32π/5. If a different value appears in a key, it might come from a different interpretation of the region, but with the stated setup, 32π/5 is the correct result.