Volume Integral Questions

Question

a) What definite integral can be used to determine the volume of a solid obtained by rotating the region bounded by x=2y^2, y=1, and x=0 about the y axis using the disks formula?

a) What definite integral can be used to determine the volume of a solid obtained by rotating the region bounded by x=2y^2, y=1, and x=0 about the y axis using the cylindrical shells formula?

Answer 1

Hey Sam, let's dive in:

(a) When using the disk method, you want to solve V= ∫π f(x)² dx (or V= ∫π f(y)² dy) as the area formula for a disk is A = π r² (where we set the function equal to r).

As we want to rotate around the y-axis, we need to add disks in the y-direction (from bottom to top). So, our integral will be with respect to y:

V= ∫π f(y)² dy with bounds from 0 to 1 (I would draw the graph to see why this is true -- it might help to know that x=2y² is equivalent to y=√(x/2) ).

So . . .

V= ∫π (2y²)² dy from 0 to 1

V= π ∫4y⁴ dy from 0 to 1

V= 4π ∫y⁴ dy from 0 to 1

V= 4π [y⁵ / 5] evaluated from 0 to 1

V = (4π/5)(1)⁵ - (4π/5)(0)⁵

V= 4π/5

(b) When using the shell method, you want to solve V= ∫2π x f(x) dx (or V= ∫2π y f(y)dy) as the area formula for a shell is A = 2π r h (where the distance from the axis to the shell is equal to x (or y) and the height of the shell is an equation involving the function f(x) (or f(y)).

As we want to rotate around the y-axis, we need to add shells in the x-direction (from left to right). So, our integral will be with respect to y.

This can also be confusing to visualize, and so I again suggest you graph the region and what a shell would look like at some arbitrary point in the range. The intersection of the graphs occurs at x=2, which will be the upper bound of the integral. Also note that the height of the shell is 1 - √(x/2) -- think "top minus buttom curve" similar to how we found the distance between two curves for area calculations.

V= ∫2π x (1 - √(x/2)) dx from 0 to 2

V= 2π ∫(x - (1/√2)x^3/2) dx from 0 to 2

V= 2π [(x²/2) - (2/5√2)x^5/2] evaluated from 0 to 2

V= 2π [ ( (2²/2) - (√2/5)2^5/2 ) - ( (0²/2) - (√2/5)0^5/2 ) ]

V= 2π ( 2 - (8/5) )

V= 2π (2/5)

V= 4π/5

Let me know if you have any questions.