Homework Help: Area under a curve (something missing from equation)

1. Dec 6, 2007

fk378

1. The problem statement, all variables and given/known data
Find the volume of the solid given the following:
y=x^2
y=0
x=1
x=2

2. Relevant equations
Shell method: 2pi(x)([f(x)] dx

3. The attempt at a solution
I shifted the function left 4 units, so I replaced x's with x+4
This makes the new functions:
y=(x+4)^2
y=0
x=-3
x=-2
Using -2 and -3 as the bounds, I have the integral of 2(pi)x (x+4)^2 dx, which is to say 2(pi)x (x^2 + 8x + 16)dx, which equals 2(pi)(x^3 + 8x^2 + 16x)
Integrating I get:
2(pi) [(x^4)/4 + (8x^3/3) + (16x^2/2)]
After evaluating it from -3 to -2, I get a negative value.

However, the right answer is supposed to be 67(pi)/6
Anyone think they can catch what's wrong?

Last edited: Dec 6, 2007
2. Dec 6, 2007

Office_Shredder

Staff Emeritus
I think this step needs a bit more justification (for example, there's no way the integral of a cubic equation will get you an x9 term)

3. Dec 6, 2007

fk378

Thanks, I just clarified my steps....where do you see an x^9 term, though?

4. Dec 6, 2007

Office_Shredder

Staff Emeritus
x^4*x^3*x^2 gives x^9.

I think you intended there to be + signs in there, and by differentiating you meant integrating.

If you fix those up, you should get the right answer

5. Dec 6, 2007

fk378

Sorry, that was actually a typo. I actually did add the terms up.

6. Dec 6, 2007

Office_Shredder

Staff Emeritus
When I added up the terms, I got the right answer. There's really not much more to say at that point

7. Dec 6, 2007

fk378

I don't think so....I keep getting a different answer than 67pi/6... Are you sure you used the same integral as I did?

8. Dec 6, 2007

Office_Shredder

Staff Emeritus
Well, doing 2(pi) [(x^4)/4 + (8x^3/3) + (16x^2/2)] and evaluating it as F(-2)-F(-3) gives the negative of the answer, but since for volume calculation you have to take the absolute value of the function (otherwise you get negative volume, and noting that as x goes from -3 to -2, x(x+4)^2 is negative, we simply need to take the negative of the answer to get what we really wanted.

I checked again, and got the same 67/6*Pi. Try typing in your calculations

9. Dec 6, 2007

fk378

Ohhh, now I'm getting the right answer in negative form. I don't understand the reasoning behind taking the absolute value (or the part where you said that (x+4)^2 is negative as x goes from -3 to -2---isn't the slope positive there?)...how do you know you just need to take the negative of it instead of thinking that you did a wrong computation?

10. Dec 6, 2007

Office_Shredder

Staff Emeritus
2*pi*r is the area of a circle, and then you replace r by f(x) to measure the distance of the function from axis of revolution. Of course, distance is positive (and 2*pi*r must be positive), so we need to throw in an absolute value sign. it's similiar to how if you're asked to find the area between the graph of x^3 and the x-axis when x is between -1 and 1, the answer isn't zero because you take the negative of the integral for negative x.

Then x(x+4)^2 is negative as (x+4)^2 is always positive, and as x is negative on the whole domain, clearly x is negative. positive*negative gives negative, so you're done.

In general, you should always put absolute value signs around your function, and then check to see when it's positive and negative. A good way to see you messed it up is if your answer comes out negative, but that won't always happen (see the x^3 example with x between -1 and 2)

11. Dec 6, 2007

Dietrick

x^2 at x=4 if you are shifting it then the equation becomes (x-4)^2. Also if you want to find a volume of a solid you need to revolve using PI*(f(x))^2.

12. Dec 6, 2007

Office_Shredder

Staff Emeritus
No, if you want to find a new variable x' so x=4 transforms to x'=0, setting x'+4=x gives x'+4=4 hence x'=0 So in the x^2 case, you get x'+4=x so (x'+4)^2=x^2

And the Pi*f^2 formula works if you're revolving around a y=constant line, otherwise you need to do 2*pi*r if you're revolving around x=constant (note f is always a function of x). You can find more online by googling washer vs. shell method

Last edited: Dec 6, 2007
13. Dec 6, 2007

Dietrick

Ok so revolved about x=4 doesnt mean moving it to x=4? Just a little unsure of the question is all.

14. Dec 6, 2007

Office_Shredder

Staff Emeritus
Well, the notation isn't great because he keeps x for the new variable. So if x=4, you want to shift it so that instead of x-4=0, you get x=0. clearly, you need to add 4 to x, so x becomes x+4