Use cylindrical shells to find the volume of the solid obtained by rotating the region bounded by y=x^2 , y=0, and x=3, about the y-axis.

Answer 1

The volume can be found by the equation: V = ∫³₀2πrh dx (where r is radius and h is height).

Here, the radius r = x, and the height h = y = x².

So, the equation becomes:

V = ∫³₀2πx(x²)dx

= 2π ∫³₀x³ dx

taking the integral → = 2π[x⁴/4]³₀

plug in x = 3:

= 2π(81/4) = 81π/2 cubic units

So the final answer is V = 81π/2 cubic units

Answer 2

Luna gave a great answer where you sum the cylindrical cross-section surface along the X-axis. I'd like to give another solution by summing the cross-sectional area of the volume along the Y-axis.

At each Y, the cross-section is a ring. To find the area of each ring, you need to know the inner radius and the outer radius. In this case, the inner radius is defined by the distance between the parabola y=x² and the Y-axis, which turns out to be √y. The outer ring, on the other hand, is bounded by the line x=3, so the outer radius is always 3. Therefore, at each Y, the area of the cross-sectional ring is

A = π(3)² - π(√y)² = π(9-y)

We then sum all of these disks from Y=0 to Y=9, which can be done via an integration. Why the bound at Y=9? That's because the volume is bounded at the intersection between the parabola y=x² and the line x=3.

V = ∫₀⁹ π(9-y) dy = π[9y-y²/2]₀⁹ = π(81-81/2) = 81π/2 cubic units