# AP Calculus Volumes

## Homework Statement

1. R is the shaded region in the 1st quadrant bounded by the graph of y=4ln(3-x), the horizontal line y=6, and the vertical line x=2
Find the volume of the solid when revolved about the horizontal line y=8

2. Let R be the region in the 1st quadrant enclosed by the graphs of f(x)=8x^3 and g(x) =sin(∏x) from x=0 to 1

Write, but do not evaluate, an integral expression for the volume of the solid generated when R is rotated about the horizontal line y=1

## Homework Equations

V=∏$\int$(outer radius$^{2}$- inner radius$^{2}$) dx from a to b

## The Attempt at a Solution

2. I thought of multiple possibilities

1. Why isn't the inner radius 8-4ln(3-x)?
2. Why Cant the outer radius be 1-sin(pix)?
Why Cant the outer radius be 1?
Why Cant the inner radius be 1-8x^3?

Last edited:

Related Calculus and Beyond Homework Help News on Phys.org
Simon Bridge
Homework Helper
You are attempting to apply the formula for the volume of a solid of revolution by the disk method.

You want the volume of the solid of revolution bounded above by two curves f(x) and g(x) inside a<x<b, about line y=c.

You should always sketch out the situation.

3. The Attempt at a Solution
... that makes the outer and inner radius the same though??

Note: vertical distance of y1=f(x) from y2=c is
|y1-y2| = |f(x)-c|=|c-f(x)|

When you evaluate these integrals, it is common to have to divide them up into regions
... since sometimes f(x) > g(x) and sometimes the other way around.

2. I thought of multiple possibilities

1. Why isn't the inner radius 8-4ln(3-x)?
2. Why Cant the outer radius be 1-sin(pix)?
Why Cant the outer radius be 1?
Why Cant the inner radius be 1-8x^3?
... sketch out the possibilities and see: what is the difference?

Last edited:
You are attempting to apply the formula for the volume of a solid of revolution by the disk method.

You want the volume of the solid of revolution bounded above by two curves f(x) and g(x) inside a<x<b, about line y=c.

You should always sketch out the situation.

... that makes the outer and inner radius the same though??

Note: vertical distance of y1=f(x) from y2=c is
|y1-y2| = |f(x)-c|=|c-f(x)|

When you evaluate these integrals, it is common to have to divide them up into regions
... since sometimes f(x) > g(x) and sometimes the other way around.

... sketch out the possibilities and see: what is the difference?
1. I thought outer radius would be that because at x=1, radius =8, and at x=0 radius is 8-4ln(3-x)
2. the functions intersect at x=0 and x=1, I thought because of that there would not be a difference in changing functions for radius.

Simon Bridge
Homework Helper
1. I thought outer radius would be that because at x=1, radius =8, and at x=0 radius is 8-4ln(3-x)
I'm not sure what you are saying there.
It helps to troubleshoot your own work if you are careful about what you say. i.e.

put
f(x)=4ln(3-x)
g(x)=6
c=8

at x=0 f(0)=4ln(3), g(0)=6, 8-6=2 and 8-4ln|3| > 2.
The two curves swap roles at x: 4ln(3-x)=6 or (3-x)^2 = e^3 x=3\pm e^(3/2) ... i.e. outside the range of interest: 0<x<2. Thus, 8-f is the upper bound, and 8-g is the lower bound, throughout the region.

2. the functions intersect at x=0 and x=1, I thought because of that there would not be a difference in changing functions for radius.
Is one of them always bigger than the other?
Think what the sketch of the two functions means for the list of questions you asked.