Disk Method

1. Jun 4, 2010

regnar

This problem does not make sense to me and plus I keep getting a different answer:
You have y=x^3 and x=y^3 these are rotated about x=-1. I put everything in terms of y and solved with respect to y. I set the problem like this:

$$\pi$$$$\int_0^1{(1 +{\sqrt[3]{y})^{2}} - (1 + y^{3})^{2}}dy$$

I foiled each function out:

$$\pi$$$$\int_0^1{(1+2{\sqrt[3]{y}+y^{2/3}) - (1+2y^{3}+y^{6})}dy$$
I'm not sure if I did this right.

Continued:

$$\pi$$$$\int_0^1{2{\sqrt[3]{y}+y^{2/3} - 2y^{3}-y^{6}}dy$$

$$\pi$$$$[2(3/4)y^{4/3}+(3/5)y^{5/3}-(1/2)y^{4}-(1/7)y^{7}]^{1}_{0}$$

$$\pi$$$$[(3/2) + (3/5) - (1/2) - (1/7)]- 0$$

$$\pi$$$$[51/35]$$

2. Jun 4, 2010

Mentallic

Sorry, before I delve into the problem, does the question ask something like:
The area between the curves y=x3 and x=y3 for $0\leq x \leq 1$ is rotated about the line x=-1. Find the volume.

3. Jun 4, 2010

Staff: Mentor

Yes, please post the exact problem statement. Inquiring minds need to know.

4. Jun 4, 2010

Mentallic

regnar, everything that you've done is correct if the question asked for the section in $0\leq x\leq 1$. But if it didn't all you have to do is change the limits

5. Jun 4, 2010

regnar

Sorry, but how would I go about changing the limits and the question exactly what put on there it gave those two function and said it rotated about x=-1.

6. Jun 4, 2010

Mentallic

I meant the limits of integration - the limits a and b in $$\int_a^bf(x)dx$$

Well as you know, the graphs y=x3 and x=y3 have an infinite domain and range, so "rotate these graphs about x=-1" doesn't tell us much. We'll assume we only need to worry about the area between the graphs in the domain $0\leq x\leq 1$, but it could just as easily be asking for the domain of $-1\leq x \leq 0$ or anything for that fact. Your answer is right, but it depends on what the question intended.

7. Jun 7, 2010

regnar

I tried what you said and I got the same answer, but here is the exact question: "Find the volume of the solid obtained by rotating the region bounded by and about the line x = -1."

8. Jun 7, 2010

Staff: Mentor

Bounded by what and what?

9. Jun 7, 2010

regnar

It doesn't specify but I'm guessing -1. I also set each function equal to each other and got 0 and 1.

10. Jun 7, 2010

Staff: Mentor

Aren't the equations y = x^3 and x = y^3 mentioned in the problem statement?

11. Jun 7, 2010

regnar

yes, I solved for x for y=x^3 and then set them equal to each other

12. Jun 7, 2010

Staff: Mentor

13. Jun 7, 2010

regnar

Im sorry, but that is the exact statement our teacher has given us.

14. Jun 7, 2010

Staff: Mentor

Is this it?
"Find the volume of the solid obtained by rotating the region bounded by y = x^3 and x = y^3 about the line x = -1."

15. Jun 7, 2010

Staff: Mentor

What do you have as the points of intersection of the two curves? There are three intersection points. I think you are working with only two of them.

16. Jun 7, 2010

regnar

I'm working with 0 and 1 but I also tried -1 and I got a negative answer unless I did that wrong which I'm not sure.

17. Jun 7, 2010

Staff: Mentor

You need two integrals for this problem - one for each of the two regions being rotated. The integral you have in post #1 is correct for the region in the first quadrant. For the region in the third quadrant, the two curves are in different places relative to each other. In other words, in the third quadrant, the outer radius is from x = y^3 and the inner radius is from x = y^(1/3).

BTW, the technique you are using is not disks - this is called the "washer" method. Each of your volume elements looks like a washer - a disk with a hole in the middle. (Not to be confused with a machine that washes clothes or dishes.)

18. Jun 7, 2010

regnar

Ohh, okay thank you very much. Also, thank you for clarifying the differences between disk and washer method.