Find Volumes of Rotated Solids in AP Calc

In summary: 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.
  • #1
jsun2015
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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=∏[itex]\int[/itex](outer radius[itex]^{2}[/itex]- inner radius[itex]^{2}[/itex]) dx from a to b

The Attempt at a Solution


1. outer radius = 8-4ln(3-x), inner radius = 8-4ln(3-x)
2. I thought of multiple possibilities

outer radius =1-sin(∏x) or 1-8x^3, inner radius =1-(8x^3) or 1-sin(∏x)

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?
 
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  • #2
You are attempting to apply the formula for the volume of a solid of revolution by the disk method.
"outer radius" and "inner radius" are kinda misleading terms.

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
1. outer radius = 8-4ln(3-x), inner radius = 8-4ln(3-x)
... 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

outer radius =1-sin(∏x) or 1-8x^3, inner radius =1-(8x^3) or 1-sin(∏x)

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?
 
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  • #3
Simon Bridge said:
You are attempting to apply the formula for the volume of a solid of revolution by the disk method.
"outer radius" and "inner radius" are kinda misleading terms.

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.
 
  • #4
jsun2015 said:
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.
 

FAQ: Find Volumes of Rotated Solids in AP Calc

1. How do you find the volume of a rotated solid in AP Calc?

To find the volume of a rotated solid in AP Calc, you will need to use the disk method or the washer method. Both methods involve integrating the cross-sectional area of the solid over the given interval. The disk method is used when the solid is rotated around the x-axis, while the washer method is used when the solid is rotated around a horizontal or vertical line other than the x-axis.

2. What are the key steps in finding the volume of a rotated solid?

The key steps in finding the volume of a rotated solid are: 1) determining which method (disk or washer) to use based on the axis of rotation, 2) setting up the integral using the appropriate cross-sectional area formula, 3) finding the limits of integration, and 4) solving the integral to find the volume.

3. Is it necessary to sketch the solid before finding its volume in AP Calc?

Yes, it is important to sketch the solid before finding its volume in AP Calc. This will help you visualize the shape of the solid and determine which method to use for finding its volume. It will also help you determine the limits of integration and check your work after solving the integral.

4. Can the volume of a rotated solid be negative?

No, the volume of a rotated solid cannot be negative. In AP Calc, the volume of a solid is always a positive value as it represents the amount of space the solid occupies.

5. What are some common mistakes to avoid when finding the volume of a rotated solid in AP Calc?

Some common mistakes to avoid when finding the volume of a rotated solid in AP Calc are: 1) using the wrong method (disk or washer) for the given axis of rotation, 2) setting up the integral with the incorrect cross-sectional area formula, 3) using the wrong limits of integration, and 4) making computational errors when solving the integral.

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