Please help me out. Thank you

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

In order to set up a problem like this, the first thing we want to know is where the two curves meet, because that is what gives us the limits of our integral. So we can set them equal to each other:

0.5x² = x

0.5x² - x = 0

x(0.5x - 1) = 0

so x = 0 and x = 2, and those are the two ends of our region. If you pick any value in between, say x = 1, the line gives y = 1 and the parabola gives y = 0.5, so the line y = x is the top of the region and the parabola y = 0.5x² is the bottom all the way across.

Now, we are rotating around the line x = 3, which is a vertical line sitting off to the right of the region. Since the axis is vertical, the shell method is the one that keeps everything in a single variable, because a thin vertical strip of the region is parallel to the axis and sweeps out a cylindrical shell when it spins. For shells we need two things, the radius and the height:

radius = 3 - x         (the distance from the strip over to the line x = 3)

height = x - 0.5x²         (top curve minus bottom curve)

Notice the radius is 3 - x and not x - 3 or just x. The strips live between x = 0 and x = 2, so they are all to the left of the axis, and 3 - x keeps the distance positive the way a radius has to be. Putting those into the shell formula 2π times radius times height, we get:

V = 2π ∫₀² (3 - x)(x - 0.5x²) dx

and that is the integral in one variable the question is asking for. If you multiply it out first it is a little friendlier to integrate:

V = 2π ∫₀² (3x - 2.5x² + 0.5x³) dx

which comes out to 2π(1.5x² - (2.5/3)x³ + 0.125x⁴) evaluated from 0 to 2, and that gives 2π(6 - 6.6667 + 2) = 8π/3, or about 8.3776.

One good way to be sure a setup like this is right is to redo it the other way and see if the two agree. If we slice horizontally instead, we solve each curve for x, so the parabola becomes x = √(2y) and the line becomes x = y, and for y between 0 and 2 the parabola is the one farther from the axis. That makes it a washer problem in y:

V = π ∫₀² [(3 - y)² - (3 - √(2y))²] dy

I worked that one out too and it also comes out to 8π/3, so the shell setup is correct. Remember, the shell version was the one asked for here since it stays in a single variable with no splitting.

So, our final answer is V = 2π ∫₀² (3 - x)(x - 0.5x²) dx, which equals 8π/3.

Answer 2

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 3

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