Find the volume of the solid whose region is bounded by y=16 and y=4x^2

I used the washer method and the shell method and I still get the wrong answers. Can anyone show me how to use the shell and washer method and still get the same answer?

Are you rotating about the x-axis or about the y-axis?

EDIT:

OK, it is VERY important to include all information, especially what you just included in "Additional Details".

The reason I asked about axis of rotation is because you mentioned using washer method and shell method, which are used when rotating about an axis. But from "Additional Details", this is NOT a volume of rotation, so we do NOT use either the disk/washer method or the shell method. This is the volume of a solid whose base is the region bounded by y = 16 and y = 4x², and whose cross sections perpendicular to the base and parallel to the x-axis are semi-circles. These are handled differently than volumes of rotation.

Since cross sections are parallel to the x-axis (or perpendicular to y-axis), we integrate with respect to y:
y = 4x² -----> x = ± 1/2 √y
Limits are y = 0 to y = 16

For each value of y, cross section perpendicular to y-axis
is a semi-circle with diameter = 1/2 √y − (−1/2 √y) = √y
and radius = 1/2 √y

Area of cross-section
= 1/2 * area of circle with radius (1/2 √y)
= 1/2 π (1/2 √y)²
= 1/8 π y

To find volume, we integrate from y = 0 to 16

V = π/8 ∫₀¹⁶ y dy
V = π/8 (1/2 y²) |₀¹⁶
V = π/16 (16² − 0²) |₀¹⁶
V = 16π

Mαthmφm

Correct, although you do not really have a choice between disk and washer methods.
Disk method is simply a specific case of washer method, when inner radius = 0
(i.e. when region to be rotated is right up against axis of rotation)

Mαthmφm

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