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Question

Write the integral in one variable to find the volume of the solid obtained by rotating the first-quadrant region bounded by y = 0.5x² and y = x about the line x = 3

Answer 1

You use the method of rings. The functions .5x^2 and x intersect at x = 0 and 2 -

x - .5x^2 = 0; x(1 - .5x) = 0; x = 0 or 1 - .5x = 0; x = 0 or x = 2;

The y difference (the height of the ring) is x - .5x^2; the radius of the ring is 3 - x (since x in [0, 2] is less than 3)

Thus, we integrate 2 π (3 - x)(x - .5x^2) from 0 to 2. Using I[a, b] for the integral from a to b and E[a,b] for the evaluation from a to b

I[0,2] 2 π (3 - x)(x - .5x^2) dx = I[0,2] 6 π x - 3 π x^2 - 2 π x^2 + π x^3 dx =

I[0,2] 6 π x - 5 π x^2+ π x^3 dx = 3 π x^2 - 5/3 π x^3+ π/4 x^4 E[0, 2] = (as the 0 terms don't contribute)

3 π * 2^2 - 5/3 π * 2^3+ π/4 * 2^4 = 12π -40/3π + 4π = 16π - 40/3π = 48/3π - 40/3π = 8/3π = 8.37758040957278

Answer 2

V = ₀∫²π(r_o² - r_i²)dy

r_o = 3 - x but since y = x, r_o = 3 - y

r_i = 3 - x but since y = 1/2x² then 2y = x² or x = √(2y) so r_i = 3 - √(2y)

So V = ₀∫²π[(3 - y)² - (3 - √(2y))²)dy

Simplifying: V = π₀∫²[(9 - 6y + y²) - (9 - 6√(2y) + 2y))dy

V = π₀∫²[y² - 8y + 6√(2y)]dy