How Do You Calculate the Volume of a Solid Rotated Around y=1?

[dlacombe13]

Homework Statement

Find the volume of the solid obtained by rotating the region bounded by the given curves about the specified line.

y=x²
x=y²
Rotated about y=1

Homework Equations

Area of cross-section (in this case, a disk) = A(x) = π(outer radius)² - π(inner radius)²
Volume = V = ∫A(x) dx

The Attempt at a Solution

[/B]
yellow line => y=x²
red line => x=y²

I converted x=y² to y=√x

The outer-radius is y=x²
The inner-radius is x=√x
The intersection points of the two graphs is (0,0) and (1,1)

So A(x) = π(x²)² - π(√x)² = π(x⁴-x)

So V = ∫A(x)dx = π ∫ [ (1/5)x⁵ - (1/2)x² ]dx integrated at (x=0 to x=1) = -3π/10 ...

However the volume can't be negative and the correct answer is 11π/30

Any help? The book shows the same graph and same intersection points, but I am getting the wrong answer.

 

[SteamKing]

Homework Statement

Find the volume of the solid obtained by rotating the region bounded by the given curves about the specified line.

y=x²
x=y²
Rotated about y=1

Homework Equations

Area of cross-section (in this case, a disk) = A(x) = π(outer radius)² - π(inner radius)²
Volume = V = ∫A(x) dx

The Attempt at a Solution

[/B]
yellow line => y=x²
red line => x=y²

I converted x=y² to y=√x

The outer-radius is y=x²
The inner-radius is x=√x
The intersection points of the two graphs is (0,0) and (1,1)

So A(x) = π(x²)² - π(√x)² = π(x⁴-x)

So V = ∫A(x)dx = π ∫ [ (1/5)x⁵ - (1/2)x² ]dx integrated at (x=0 to x=1) = -3π/10 ...

However the volume can't be negative and the correct answer is 11π/30

Any help? The book shows the same graph and same intersection points, but I am getting the wrong answer.

You want to use the shell method here:

https://en.wikipedia.org/wiki/Shell_integration

 

[dlacombe13]

Thank you for you're reference, I read it and sort of get how it works. However this chapter I am in does not show me that method yet, and only shows the disk/ washer method that I attempted, so it must be solvable using that method. In fact, there is an almost identical problem using y=x² and y=x revolved around the x-axis (y=0) that uses this method and it works. Any help trying to solve it using this method?

 

[dlacombe13]

Turns out I found out what I was doing wrong. I had to subtract 1 (y=1) from each of the outer and inner radius. So the inner radius is π(1-x²)² and the outer radius is π(1-√x)². When I integrated the difference of these two at 1 and 0 I got 11π/30.

 

[SammyS]

Turns out I found out what I was doing wrong. I had to subtract 1 (y=1) from each of the outer and inner radius. So the inner radius is π(1-x²)² and the outer radius is π(1-√x)². When I integrated the difference of these two at 1 and 0 I got 11π/30.

I think you have something close to the right idea.

Actually, the radii are: $\ 1-x^2 \$ and $\ 1-\sqrt{x\,}\$. However, y = x² is the lower curve (farther away from y = 1), so $\ 1-x^2 \$ is the outer radius.