# Volume of the solid

1. Mar 2, 2005

Find the volume of the solid obtained by rotating the region bounded by the given curves about the specified axis.

y=x^6
y=1

here's what i done:

$$x=(y)^{1/6}$$
$$A(y) = pi(4-(y)^{1/6})^2$$

$$\int_0^{1} pi(4-y^{1/6})^2 dy$$

anyone know what i have done wrong? this section really confuses me.

2. Mar 2, 2005

### Davorak

In which plane are the circles? x-y, x-z,y-z? The radius of the circle is defined by what?

3. Mar 2, 2005

x-y

"The radius of the circle is defined by what?"

are you asking me or telling me to post more about the question? cause that's the whole question.

4. Mar 2, 2005

### Davorak

If the equations are defined in the x-y plane then the circles have to be in the x-z or x-y plane. When you rotate around y=4 you are rotating in to the z axis.

y=4 and y = 1 are horizontal lines in the y vs x plane. Try visualizing the problem keeping this in mind.

Edit:
Does that help?

Last edited: Mar 2, 2005
5. Mar 2, 2005

### HallsofIvy

Staff Emeritus
The point of the question about the radius of the circles (formed when rotating the figure) is that since you are rotating around the line y= 4, the radii are in the y direction, not x!

6. Mar 2, 2005

### BobG

You're rotating around the wrong axis. You're rotating around x=0, not y=4. Try the washer method instead of the disc method and rotate around y=4.

7. Mar 3, 2005

this section i'm doing is all about the disc method, we havent learned the washer method yet(if possible), so can you please help me with the disc method. i see that i was rotating about the x=0 axis, how would i change it to rotate it around the y=4 axis?

i'm thinking....

$$\int_0^{1} pi*(4-x^6)^2 dx$$

8. Mar 3, 2005

### HallsofIvy

Staff Emeritus
That is exactly what every one has been doing! Rotating around the y= 4 axis just means that the radius of each disk is measured by y- 4, not x.

9. Mar 3, 2005

hmm then what am i doing wrong? after integrating that function i get 1359/91*pi. only thing that i can think of getting incorrectly is the bounds which looks correct because i plugged it into my calculator.

10. Mar 3, 2005

### Jameson

Ok, first of all, the bounds. You set the two equations equal to each other to and solve.

$$x^6=1$$ The answers to this are (-1,1)...QED: These are your bounds.

The volume using the disc\washer method is found by:

$$V = \pi\int_{a}^{b}R^2-r^2dx$$ (dx in this case!)

Ok. So, since you are revolving about the line y=4 the outer radius becomes (4-x^6)^2 and the inner radius is (4-1)^2.

Alright, finally lets set it up.

$$V = \pi\int_{-1}^{1}(4-x^6)^2-(3)^2 dx$$

I'll leave to you to figure that out.

11. Mar 3, 2005

awesome, thanks for the help. the only thing that i dont get is how you got an inner radius of 3?

12. Mar 3, 2005

### Jameson

4-1=3... the axis of rotation is y=4, so the area you're rotating is 4-1