Volume of solid of revolution - y axis.

In summary, the conversation discusses finding the volume of a solid of revolution by rotating the area between the x-axis and the function f(x) = 1 - cosx around the y-axis. The solution involves using the formula V = ∫ 2∏x f(x) dx, with the limits of 0 and 2∏ and the function x(1-cos(x)). After correcting a mistake in the integration by parts, the final answer is found to be 2∏^3 - 4∏ + 2.
  • #1
Mutaja
239
0

Homework Statement



Find the volume of the solid of revolution when we rotate the area limited by the x-axis and the function f(x) = 1 - cosx where x e [0, 2∏] once around the y-axis?

The Attempt at a Solution



In my notes I have the following equation:

V = ∫ 2∏x f(x) dx

If I put in my limits (upper limit 2∏, lower limit 0) and my function I get the following:

V = 2∏ ∫x(1-cos(x)) dx

V = 2∏ ∫x - xcos(x) dx

V = 2∏[[itex]\frac{x^2}{2}[/itex] - (xsin(x)+cos(x))]

V = 2∏ [[itex]\frac{x^2}{2}[/itex] - (∏sin(∏) + cos(∏)] - 2∏ [[itex]\frac{0^2}{2}[/itex] - (0sin(0) + cos(0)]

Since ∏ sin(∏) = 0, cos(∏) = -1 , 0sin(0) = 0 and cos(0) = 1 I get the following:

V = 2∏ ([itex]\frac{∏^2}{2}[/itex]) - 2∏ + 1

V = ##2∏^3## - 4∏ + 2

Is this correct? Am I using the correct formulas/equations?

Please let me know if there is something I need to explain better. Any help and guiding is massively appreciated. Thanks.
 
Last edited:
Physics news on Phys.org
  • #2
Mutaja said:
V = 2∏ ∫x - xcos(x) dx

V = 2∏[[itex]\frac{x^2}{2}[/itex] - xsin(x)+cos(x)]

Check your integration by parts: should be -cos(x).
 
  • #3
PeroK said:
Check your integration by parts: should be -cos(x).

Sloppy mistake by me, there should of course be parenthesis around that equation.

V = 2∏[[itex]\frac{x^2}{2}[/itex] - (xsin(x)+cos(x))]

That slipped past me when I was writing off of my notes - therefore unless I again have overlooked something (which I've double checked I haven't), that mistake was a one off - I've solved the rest of the problem as if there were parenthesis around (xsin(x)+cos(x)). Therefore I get - (-1) -> +1 in my answer which I later multiply by 2.

Does it look ok besides that error?

Thanks for your input!
 

Related to Volume of solid of revolution - y axis.

1. What is the definition of "volume of solid of revolution - y axis"?

The volume of a solid of revolution is the amount of space that is enclosed by a three-dimensional figure created by rotating a two-dimensional shape around an axis. In the case of the y-axis, the shape is rotated around the y-axis to create the solid.

2. How is the volume of a solid of revolution calculated?

The volume of a solid of revolution can be calculated using the disk or washer method. The disk method involves slicing the solid into thin discs, calculating the volume of each disc, and then summing up all the volumes. The washer method is similar, but takes into account the hollow center of the solid.

3. What types of shapes can be used to create a solid of revolution around the y-axis?

Any two-dimensional shape that is symmetrical around the y-axis can be used to create a solid of revolution. This includes circles, rectangles, triangles, and even more complex shapes such as parabolas or ellipses.

4. How does the shape of the two-dimensional figure affect the volume of the solid of revolution?

The shape of the two-dimensional figure used to create the solid of revolution will affect its volume. For example, a larger radius of rotation will result in a larger volume, while a smaller radius will result in a smaller volume. Also, the shape of the two-dimensional figure will determine the cross-sectional area of the solid, which will impact the overall volume.

5. What real-world applications involve calculating the volume of a solid of revolution around the y-axis?

The volume of a solid of revolution around the y-axis is commonly used in engineering, architecture, and physics. For example, it can be used to calculate the volume of a water tower, the capacity of a cylindrical tank, or the volume of a cone-shaped rocket nose cone.

Similar threads

  • Calculus and Beyond Homework Help
Replies
3
Views
1K
  • Calculus and Beyond Homework Help
Replies
2
Views
525
  • Calculus and Beyond Homework Help
Replies
2
Views
910
  • Calculus and Beyond Homework Help
Replies
3
Views
403
  • Calculus and Beyond Homework Help
Replies
9
Views
1K
  • Calculus and Beyond Homework Help
Replies
2
Views
510
  • Calculus and Beyond Homework Help
Replies
27
Views
2K
  • Calculus and Beyond Homework Help
Replies
9
Views
397
  • Calculus and Beyond Homework Help
Replies
18
Views
1K
  • Calculus and Beyond Homework Help
Replies
14
Views
756
Back
Top