Calculate Volume of Solid Rotated about y=4

[ProBasket]

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
about y=4

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.

 

[Davorak]

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

 

[ProBasket]

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.

 

[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 into 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?

 

[HallsofIvy]

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!

 

[BobG]

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
about y=4

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.

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.

 

[ProBasket]

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.

this section I'm doing is all about the disc method, we haven't 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...

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

 

[HallsofIvy]

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.

 

[ProBasket]

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.

 

[Jameson]

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
about y=4

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.

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!)
R is the outer radius and r is the inner radius.

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 let's set it up.

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

I'll leave to you to figure that out.

 

[ProBasket]

awesome, thanks for the help. the only thing that i don't get is how you got an inner radius of 3?

 

[Jameson]

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