Calculus 2 Need help

Question

Use the shell method to find the volume of the solid generated by revolving the regions bounded by the curves and lines about the​ y-axis.

y=72 - x2, y =7x2, x = 0 for x ≥ 0

The volume of the solid is __

cubic units.

Answer 1

Hi Joseph,

Without even drawing a picture, you can determine the intersection points of the curves by setting them equal to each other and solving for x.

72 - x² = 7x²

72 = 8x²

x² = 9

|x| = 3, or,

x = -3 and x = 3

The problem states that we only care about x > 0, so I will disregard the x = -3.

If you draw a picture of the two curves from x = 0 to x = 3, you will find that 72 - x² is above 7x² the entire time.

Rotate this region around the y-axis. You get a weird looking UFO shape that looks prone to tipping over.

A cylindrical shell is essentially a piece of paper that's been rolled to resemble a soup can with no top or bottom. The unrolled piece of paper resembles a rectangular prism, which has volume equal to its length * width * height.

The length of the sheet of paper is the same as the circumference of the cylinder (2π*r)

The height of the sheet of paper depends on how it's going to sit inside your region. We'll give height the letter "h".

The width of the sheet of paper is basically its thickness. We'll call it "t"

So the volume of one shell has the following formula derived from your rectangular prism calculation

V = 2π*r*h*t

We'll use this formula to determine our "fundamental cylindrical shell". This is the basic shell formula that will model every shell in the region.

Since we're rotating around the y-axis, the radius of your cylinder will start there and move horizontally to the right until it hits one of the curves. This means the radius is just the x-value of the curve. r = x

The height of the cylinder will be the vertical length traveled between the two curves. Since 72 - x² is above 7x² throughout the entire region, the formula for the height will be:

h = (72 - x²) - (7x²)

h = 72 - 8x²

The thickness of our piece of paper is infinitesimally small. It will always either be dy or dx. If you imagine rolling up your piece of paper and placing it vertically in your region, you'd be able to take scissors and cut vertically through the wall. That cut could expose a wall that has a thickness in the left to right direction. Since the wall thickness is along the x-axis, we'll be using dx for this problem.

On the flip side, if we had rotated our region around the x-axis, all your rolled up pieces of paper would be on their side. You'd need to take scissors horizontally across the cylinder to open it up, and the wall would have thickness in the up and down direction. That would necessitate using dy as your thickness.

Updating our volume formula with our specific values, we get:

V = 2π*r*h*t

V = 2π*x*(72 - 8x²)*dx

= (144π*x - 16πx³)*dx

This will give us the volume of one cylindrical shell at any point x we decide to choose. That doesn't help us a whole lot, though, since we want the volume of the entire region. The region encompasses x-values from 0 to 3 as determined above, so let's integrate this formula with those bounds.

3

∫(144π*x - 16πx³)*dx

0

3

= 72π*x² - 4πx⁴ |

0

= 72π*(9) - 4π(81)

= 648π - 324π

= 324π

= ~1018 cubic units

Hope this helps.

Answer 2

72 - x² = 7x²; x² = 9; x = 3.

By shell method volume v = 2π∫₀³(72 - x² - 7x²)xdx = 2π(36x² - 2x⁴)₀³ = 2π(36·9 - 162) = 324π