Question

CALCULUS: Which of the following represents the volume of the solid formed by revolving the region bounded by the graphs of y =
`x^(3)`, y = 1, and x = 3, about the line x = 3?

A. π * [27,1]∫ (3-∛y)² dy
B. π * ∛
`\int\limits^3_1 {(3-\sqrt[3]{y})^(2) } \, dy`
C. None of these

Answer 1

π * [27,1]∫ (3-∛y)² dy

Explanation:

Answer 2

Disk method it is, since both given options integrate with respect to y.

First find where the boundaries of the region intersect.

• y = x ³ and y = 1 intersect at (1, 1)

• y = x ³ and x = 3 intersect at (3, 27)

• y = 1 and x = 3 intersect at (1, 3)

So the bounded region is the set of points

(x, y)

But since we're integrating with respect to y, rewrite this set so that y has constant limits:

∛y ≤ x ≤ 3 and 1 ≤ y ≤ 27

Pick some point y in the interval [1, 27] and construct a disk centered at the given axis of revolution (x = 3). Such a disk will have radius 3 - ∛y because x = ∛y is the horizontal distance from the y-axis to the curve y = x ³ and x = 3 is itself 3 units away from the y-axis. Its height will be some small change in y, call it ∆y. Then the volume of this disk is

π (3 - ∛y )² ∆y

Now do the same thing for every y in [1, 27] - infinitely many of them! - and make ∆y very small, such that ∆y → dy. The volume of the solid is the sum total of the volumes of these infinitely many disks, given by the integral

`\displaystyle\pi\int_1^(27)\left(3-\sqrt[3]{y}\right)^2\,\mathrm dy`

and so the answer is A (assuming the limits of integration are listed from upper to lower).