Find the volume of the solid generated by rotating the region bounded

In summary: So the constant 2 is not the issue, but the 1/3 is.In summary, we are using the washer method to find the volume of a solid generated by rotating the region bounded by the x-axis, the curve y=3x^4, and the lines x=1 and x=-1, with the axis of rotation being the y-axis. We set up the integral using the washer method and use the symmetry of the function to simplify the expression. However, we should not multiply by 2, and instead use only one half of the region to find the volume. Additionally, we should use the substitution u=y/3 to correctly evaluate the integral.
  • #1
jorgegalvan93
10
0

Homework Statement



Find the volume of the solid generated by rotating the region bounded by the x-axis, the curve y=3x^4, and the lines x=1 and x= -1. The axis of rotation is the y-axis.


Homework Equations



Washers method: V=∏∫ [(R)^2 - (r)^2]dr
x = (y/3)^(1/4)

The Attempt at a Solution



So for the following problem, I used washers method to solve for the volume of the solid. R representing the outer radius, and r representing the inner radius (hole).
Because my washer is perpendicular do the y-axis, I will set my expression with respect to y… dy
I will use symmetry, so my R is 1, and r is a vertical distance so it is x, which is equal to (y/3)^(1/4)

V=∏∫ [(1)^2 - ((y/3)^(1/4))^2]dy , where the limits of integration are a = 0 and b = 3
V = 2∏∫ [(1) - √(y/3)]dy (step where problem arises)
V = 2∏[(y - 2/3(y/3)^(3/2)] from 0 to 3
V = 2∏[3 - 2/3(3/3)^(3/2)]
V = 2∏[3 - 2/3(1)]
V = 2∏[3 -(2/3)] … V = 2∏(7/3)….

V = 14∏/3

My professor said that on the second step performed, I should have done a substitution… u = y/3. I don't think that's correct though, because I would still have a 1.
Can anyone give me a hand with this problem?

Thank you!
 
Physics news on Phys.org
  • #2
jorgegalvan93 said:

Homework Statement



Find the volume of the solid generated by rotating the region bounded by the x-axis, the curve y=3x^4, and the lines x=1 and x= -1. The axis of rotation is the y-axis.


Homework Equations



Washers method: V=∏∫ [(R)^2 - (r)^2]dr
x = (y/3)^(1/4)

The Attempt at a Solution



So for the following problem, I used washers method to solve for the volume of the solid. R representing the outer radius, and r representing the inner radius (hole).
Because my washer is perpendicular do the y-axis, I will set my expression with respect to y… dy
I will use symmetry, so my R is 1, and r is a vertical distance so it is x, which is equal to (y/3)^(1/4)

V=∏∫ [(1)^2 - ((y/3)^(1/4))^2]dy , where the limits of integration are a = 0 and b = 3
V = 2∏∫ [(1) - √(y/3)]dy (step where problem arises)
V = 2∏[(y - 2/3(y/3)^(3/2)] from 0 to 3
V = 2∏[3 - 2/3(3/3)^(3/2)]
V = 2∏[3 - 2/3(1)]
V = 2∏[3 -(2/3)] … V = 2∏(7/3)….

V = 14∏/3

My professor said that on the second step performed, I should have done a substitution… u = y/3. I don't think that's correct though, because I would still have a 1.
Can anyone give me a hand with this problem?

Thank you!
You are combining some aspects of the washer method and some aspects of the shell method.

The washer method should look like
[itex]\displaystyle \pi\int_{a}^{b}\,(f(y))^2-(g(y))^2\,dy\ .[/itex]​
 
  • #3
That's the method I used. Where f(y) = 1 and g(y) = (y/3)^(1/4)
I have a 2 next to ∏ because I am using the symmetry of the function. I thought it would be much simpler that way. But I suppose I can find the volume without using symmetry. By doing that, f(y) = 2 and g(y) = 2x?
I'm confused as to what g(y) would have to be in this instance.
 
  • #4
jorgegalvan93 said:
That's the method I used. Where f(y) = 1 and g(y) = (y/3)^(1/4)
I have a 2 next to ∏ because I am using the symmetry of the function. I thought it would be much simpler that way. But I suppose I can find the volume without using symmetry. By doing that, f(y) = 2 and g(y) = 2x?
I'm confused as to what g(y) would have to be in this instance.
Do not multiply ny 2. Only use the right half or the left half of the region. Either will "sweep out" the entire volume that you're interested in.

Also, I agree with your professor. The 1 is not the issue.

I integrating [itex]\displaystyle \ \ \sqrt{\frac{y}{3}}\,, \ [/itex] you have that the anti-derivative of that is [itex]\displaystyle \ \ \frac{2}{3}\left(\frac{y}{3}\right)^{3/2}\ .[/itex]

Now, if you check that result by taking the derivative, you get [itex]\displaystyle \ \ \frac{2}{3}\frac{3}{2}\left(\frac{y}{3}\right)^{1/2}\left(\frac{1}{3}\right) \ .\ [/itex] The last 1/3 coming from the chain rule.
 

FAQ: Find the volume of the solid generated by rotating the region bounded

What is the definition of volume?

Volume is the amount of space occupied by a solid, liquid, or gas.

How do you find the volume of a solid?

To find the volume of a solid object, you need to measure the length, width, and height of the object and then use the formula V = lwh, where V represents volume, l represents length, w represents width, and h represents height.

What does it mean to rotate a region?

Rotating a region means to move it around a fixed point, creating a three-dimensional shape. In this context, we are rotating a two-dimensional region to create a solid object.

What is the difference between rotating a region and revolving a solid?

Rotating a region involves moving a two-dimensional shape around a fixed point to create a solid, while revolving a solid involves spinning a three-dimensional object around an axis to create a different three-dimensional shape.

How do you find the volume of a solid generated by rotating a region?

To find the volume of a solid generated by rotating a region, you need to use the formula V = ∫(πr^2)dx, where V represents volume, ∫ represents the integral symbol, r represents the distance from the axis of rotation, and dx represents the small width of the region being rotated.

Back
Top