volume by washer method when f (x) below x axis

[yamanaha]

Find volume by revolution. Shell method not taught...so this needs to be done as washer. I understand most of these but get confused when the function is below the x axis. Usually the distance from the x axis to the top of the area is defined by distance by f (x)....but this is flipped. So i usually add the distance from the line of revolution to x axis. I am thinking this way at least so please help. If I have y=x^2-9, y=0. About the line y=1. . What differs when its f (x) is below x axis like this??

 

[stapel]

Find volume by revolution. Shell method not taught...so this needs to be done as washer. I understand most of these but get confused when the function is below the x axis. Usually the distance from the x axis to the top of the area is defined by distance by f (x)....but this is flipped. So i usually add the distance from the line of revolution to x axis. I am thinking this way at least so please help. If I have y=x^2-9, y=0. About the line y=1. . What differs when its f (x) is below x axis like this??

If the function is below the x-axis, then flip it. If f(x) is negative, then use -f(x).

 

[Quaid]

get confused when the function is below the x axis

The absolute value of those negative function values (plus 1 more each) are the outer radii; they get squared, so there is nothing different from calculating the volume using y=9-x^2 about y=-1. (Turn your sketch upside down, and compare.)

In other words, set up the integral in the same way that you would, if the area being revolved were to lie above the x-axis, instead.

If you're still confused, let us know. We can draw you a picture.

:cool:

 

[yamanaha]

I should have said for getting R...the outer radius. I see what you mean about -f (x) being the positive..and that helped. I also see that squaring a neg or pos will get the same value...the problem is this one is displaced bc the axis of rotation is y=1. So my confusion is...you add or subtract the 1 to f (x) before you square it so wouldnt that give me 2 diff answers depending on if I negate the function first or not? So I guess I have to negate f (x) and add 1..then square it for R?

I get the formula for washer is (outer radius - inner radius )integrated times pi. My only difficulty is finding outer radius always correctly.

Would it be outer radius = -(x^2-9)+1

Still confused. help please!

 

[yamanaha]

I get that r..the inner radius 1, is outer radius -x^2+10?

 

[yamanaha]

Those negative function values are the outer radius; they get squared, so there is nothing different from calculating the volume using y=9-x^2 about y=-1. (Turn your sketch upside down, and compare.)

In other words, set up the integral in the same way that you would, if the area being revolved were to lie above the x-axis, instead.




That does not sound right.

Washers have a hole in the center, so you ought to be subtracting the volume of the hole, not adding it.

If you're still confused, let us know. We can draw you a picture.

:cool:

still confused

 

[Quaid]

still confused

Hi yamanaha:

Please excuse me; I misunderstood part of your first post, and I botched what I was trying to say, anyway. (I edited my first reply.)

Yes, I understand what you said; you normally add the inner radius to the length which comes from the revolved region, to get the outer radius.

With this exercise (where y values are negative), we could think of the radii in terms of absolute values -- while looking at a sketch.

Here's my sketch (axis scales not proportional).



Let's think! We know that the outer radius needs to be 10 units, when x = 0 (i.e., the distance from the region's y-intercept to the line of revolution).

We also know that the outer radius needs to be 1 unit, when x = 3 (i.e., the distance from the x-axis to the line of revolution).

Let R represent the outer radius, and let r be the inner radius.

When x = 0, y = -9. We want R = |-10| there, so we need to subtract 1 from y.

When x = 3, y = 0. We want R = |-1| there, so again we need to subtract 1 from y.

Therefore, R must be |y - 1|. (You already know that r = 1.)

Because R gets squared, we may drop the absolute value symbols.

R = x^2 - 10

Volume = 2π ∫(R^2 - r^2) dx where x goes from 0 to 3.

Hope I got it right, this time.

I get a volume of 1040½ cubic units (rounded). You too?

:cool:

 

[Quaid]

PS: I think the lesson to learn is this. When the region being rotated lies below the x-axis, then we need to subtract the inner radius (instead of adding it), to get R. If we forget this, then we can remember by drawing a sketch and reasoning it out -- using some actual values to check our reasoning.

 

[yamanaha]

That makes so much sense Quaid when you use ordered pairs and compare to what I can easily see needs to be the distance. Not sure why I never thought of that but I am sure glad you did!! Thank you so much. Woohoo!! Successfully understand it finally. The absolute value and flipping the function concept helped too.

and yes I got 332.2 pi or 1040.5