Finding Volume of Rotated Region Bounded by Equations

[Jovy]

Homework Statement

Find the volume of the solid generated by revolving the region bounded by the graphs of the equations about the line y =4.
$$y=\frac 3 {1+x},~ y=0,~ x=0,~x=3$$

Homework Equations

$$V= \int_a^b ([R(x)]^2-[r(x)]^2)dx$$

The Attempt at a Solution

I understand how to use the equation, but I don't know how to find all the components needed to plug into the equation. R(x) and r(x) are both the inner and outer radius, however, I don't understand why R(x)=4 and $r(x)=4-\frac 3 {1+x}$. I understand that (a,b) are (0,3) since that is what x equals. I'm just having trouble understanding the radius.

This website showed how to solve the problem, but it doesn't really explain each step.
http://www.calcchat.com/book/Calculus-ETF-6e/7/2/17/

 

[Student100]

Homework Statement

Find the volume of the solid generated by revolving the region bounded by the graphs of the equations about the line y =4.
$$y=\frac 3 {1+x},~ y=0,~ x=0,~x=3$$

Homework Equations

$$V= \int_a^b ([R(x)]^2-[r(x)]^2)dx$$

The Attempt at a Solution

I understand how to use the equation, but I don't know how to find all the components needed to plug into the equation. R(x) and r(x) are both the inner and outer radius, however, I don't understand why R(x)=4 and $r(x)=4-\frac 3 {1+x}$. I understand that (a,b) are (0,3) since that is what x equals. I'm just having trouble understanding the radius.

This website showed how to solve the problem, but it doesn't really explain each step.
http://www.calcchat.com/book/Calculus-ETF-6e/7/2/17/

Did you draw a picture? That's most helpful for these problems. Try to actually draw the solid after rotation.

So the solid you're trying to find the volume of is rotated about the line y =4. Now, what you're trying to determine is an inner and outer radius from the axis of rotation. So the outer radius is the bottom function, in this case it's just the x-axis (the line y=0). The distance of which is a constant 4 from the axis of rotation. The inner radius varies with the function provided. So it changes as you move along the path of integration. Try to set up a picture now, draw the resultant solid, then try to see if you can visualize the radii.

 

[Jovy]

Did you draw a picture? That's most helpful for these problems. Try to actually draw the solid after rotation.

So the solid you're trying to find the volume of is rotated about the line y =4. Now, what you're trying to determine is an inner and outer radius from the axis of rotation. So the outer radius is the bottom function, in this case it's just the x-axis (the line y=0). The distance of which is a constant 4 from the axis of rotation. The inner radius varies with the function provided. So it changes as you move along the path of integration. Try to set up a picture now, draw the resultant solid, then try to see if you can visualize the radii.

Here is my drawing, I'm not sure I drew it correctly. The inner radius is the hole, correct? So wouldn't the inner radius be what is in between the bottom drawing and the top?

 

[Student100]

Here is my drawing, I'm not sure I drew it correctly. The inner radius is the hole, correct? So wouldn't the inner radius be what is in between the bottom drawing and the top? View attachment 113477

So for the artistically impaired (like me), I would do this..

Do you see how it's mirrored about the axis of rotation?

 

[Jovy]

I see where the outer radius, however, I don't understand where the inner radius would be.

 

[Student100]

I see where the outer radius, however, I don't understand where the inner radius would be.
View attachment 113484

Draw a few lines from the axis of rotation to curve $\frac{3}{1+x}$. You can also then see why it needs to be $4- \frac{3}{1+x}$ then.

Look at the case x=0 and x=3 specifically.

 

[Jovy]

Draw a few lines from the axis of rotation to curve $\frac{3}{1+x}$. You can also then see why it needs to be $4- \frac{3}{1+x}$ then.

Look at the case x=0 and x=3 specifically.

Is it because the inner radius is below y=4 and what is below that is $\frac{3}{1+x}$. So it would be $4- \frac{3}{1+x}$?

 

[Student100]

Is it because the inner radius is below y=4 and what is below that is $\frac{3}{1+x}$. So it would be $4- \frac{3}{1+x}$?View attachment 113487

The inner radius is the curve that forms the part of the solid closest to the axis of rotation, while the outer radius is the one furthest away. It's not easy for me to put into words..

Try reading: http://tutorial.math.lamar.edu/Classes/CalcI/VolumeWithRings.aspx see if that helps with their better pictures. Pay close attention to example 3.

 

[LCKurtz]

Is it because the inner radius is below y=4 and what is below that is $\frac{3}{1+x}$. So it would be $4- \frac{3}{1+x}$?View attachment 113487

You have shaded an incorrect area that is being rotated. Shade the region under $y=\frac 3 {1+x}$ above the $x$ axis between $x=0$ and $x=3$. If you are thinking about the wrong area no wonder you don't get the radius correct.