# Homework Help: Disks, Shells, Washers- Calculus AB/BC

1. May 22, 2013

### Justabeginner

1. The problem statement, all variables and given/known data
Find the area bounded by the curves $f(x)= x^3 + x^2$ and $g(x)= 2x^2 + 2x$.

2. Relevant equations
\int ((f(x)-g(x))^2)\, dx and \int ((g(x)-f(x))^2)\, dx

3. The attempt at a solution
I found out the intersection of f(x) and g(x) first. I think there are three points: -1, 0, and 2. I then set the a and b values for those integrals -1 and 0, and 0 and 2 respectively. I got -13pi/105 as the result of the first integral, and -11104pi/105 as the result of the second integral. Adding them up, I got 11117pi/105 and this is incorrect apparently. I would appreciate some guidance on if my technique is correct or not, or if I've just messed up a calculation. Thank you so much. :)

Last edited: May 22, 2013
2. May 22, 2013

### clamtrox

How do you calculate the area between two curves? It should be A = ∫|f(x)-g(x)| dx, right? How, then, can you possibly get pi's in your answer?

3. May 22, 2013

### Justabeginner

I did not realize that. Thank you for the clarification. I'll solve it in that way now.

I solved it in that method, and got 59/12 as my final answer, but it's not correct. o.0 Did I do something wrong?

4. May 22, 2013

### clamtrox

Yes you did, as is evident from the fact that your answer is wrong. However, it's terribly difficult to guess what your error may be, without seeing your working.

5. May 22, 2013

### Justabeginner

Well my work involved solving the integral of f(x)-g(x) with a and b being -1 and 0 respectively and adding it to the integral of g(x)- f(x) with a and b being 0 and 2 respectively.

Oh wait, I didn't do the absolute value of that, just as you showed in the first post. Can you explain why you're using the absolute value? Because in other problems solved, since part of the equation was under the x-axis negative area was involved.

Doing it with the absolute value, I now get:

1/4* x^4 + 1/3* x^3 + x^2 (with a being -1 and be being 0)
11/12 - 0= 11/12

For the second part with a being 0 and b being 2, I get:

32/3- 0 = 32/3

This is STILL not right. I'm a bit frustrated, as I've literally done this about fifteen times, and I still can't get it. :/

6. May 22, 2013

### CAF123

I think there are some sign errors in the line *

I get something different - could you have a possible sign error again?

7. May 22, 2013

### Justabeginner

I have returned with the right answer! And a better understanding of the approach. :D

Yes, CAF123, thank you- I did have a sign error.

For the left hand side I come out with 5/12 and for the right hand side, 8/3. Adding the two together, I get 37/12, which is correct!

And thank you Clamtrox for helping me with the main equation- I now see what my main mistake was!

I should really not get flustered and pay closer attention to my signs next time. -.-

8. May 22, 2013

### clamtrox

Absolute value is there because there's no such thing as negative area. You should order the functions so that you subtract smaller from the larger value.

9. May 22, 2013

### Staff: Mentor

You don't need the absolute value if you set each integral up correctly, and in some cases, using the absolute value will give you the wrong answer. If f(x) ≥ g(x) on [a, b], then the integrand should be f(x) - g(x). If f(x) ≤ g(x) on [b, c], then the integrand should be g(x) - f(x).

10. May 22, 2013

### CAF123

In this case on $[-1,0], f(x) \geq g(x)$, however $\int_{-1}^0 f - g \,\,dx < 0$(because the majority of the region enclosed is below x axis), so in this case an abs value sign is needed.

11. May 22, 2013

### Staff: Mentor

No, that's incorrect. If f(x) ≥ g(x) on an interval [a, b], then f(x) - g(x) ≥ 0 on the interval, so $\int_a^b f(x) - g(x)~dx \geq 0.$

As a simple example of how this works, consider -2 and -4. -2 > -4 and -2 - (-4) = +2 > 0.

12. May 22, 2013

### Justabeginner

Yes, but in certain texts, I've seen the person solving the problem, say the area is negative simply because the area extends below the x-axis. o.0 Can you explain why that holds true in that case, or was it an error on their part?

13. May 22, 2013

### Justabeginner

That really cleared up some things. Thank you :)

14. May 22, 2013

### CAF123

Yes, sorry Mark.

15. May 22, 2013

### Staff: Mentor

Error on their part.
The value of an integral can be negative; e.g. $\int_{-1}^1 x^2 - 1~ dx$,
but the area of a geometric region is always nonnegative.

If you're asked to find the area of the region that is bounded above by the x-axis and below by the graph of y = x2 - 1, you should get a positive number.

16. May 22, 2013

### Justabeginner

I understand. Thank you so much for clarifying.