How to Find Volume of Solid of Revolution for a Cycloid Curve?

[ronaldor9]

Homework Statement

The region above the curve y = x² − 6x + 8 and below the x-axis is revolved around the line x = 1. Find the volume of the resulting solid.

Homework Equations

$$\int 2\pi x f(x)$$

The Attempt at a Solution

$$\int_2^4 2 \pi (x-1)(x^2-6x+8)= -\frac{16 \pi}{3}$$

Homework Statement

checking the solution the answer is correct but it is negative, why?

Second question:

Homework Statement

The region under the arch of the cycloid $$x = a - a \sin\theta, \quad y = a - a \cos\theta, \quad 0 \leq \theta \leq 2\pi$$ is revolved around the x-axis. Find the volume of the solid of revolution produced.

Here I am totally lost since the forumla for the solid of revolution around the x-axis is only in terms of x and not a parametrically defined curve. How should i attack this problem?

 

[Mark44]

For your first problem, the height of your typical area element (which will be revolved around the line x = 1 to produce a volume element) is 0 - (x² -6x + 8). What you have is the opposite in sign, thereby producing an answer whose sign is off.

I can't tell what you have for the parameteric equations in your second problem, as some of the symbols are showing up as squares.

 

[Chrisas]

If the weird symbols are supposed to represent the parameter, and you haven't covered methods that involve a parametric representation, then I would recommend trying some algebra/trig to eliminate the parameter from the two equations by combining them into one equation. It's hard to tell if that would work until the format of the problem is fixed.

 

[ronaldor9]

Hey I fixed the problem by formating it in TeX. I tried removing the parameter but I think it is not possible?

 

[Mark44]

x - a = -asin t, y - a = -acos t

Now, square both sides of each equation, and get a new equation by adding the left sides together and the right sides together. Voila, no more parameter.