Calculating Surface Area of a Revolution Rotated About the Y-Axis

In summary, the surface area for the given equation rotated about the Y-axis is 94925010.28. This is calculated by integrating 2πf(y)√1+ (dx/dy)^2 from 0 to 216 and multiplying the result by 2 and then by 2π, where f(y) = 3y^(4/3) - 3/32y^(2/3) and dx/dy = 4y^(1/3) + 1/16y^(-1/3).
  • #1
deerhake.11
7
0
(my first dealings with latex.. so bare with me if this looks a little messed up at first :rolleyes: )

Homework Statement


Find the surface area for the equation:
[tex]x = 3y^{4/3} - \frac{3}{32}y^{2/3}[/tex]

with bounds [tex]-216 \leq y \leq 216[/tex]

rotated about the Y-axis.

Homework Equations



[tex]\int^a_b 2\pi f(y) \sqrt{1+(\frac{dx}{dy})^2}[/tex]

The Attempt at a Solution



well... going with that equation i get to this point:

[tex]2\pi \int^{216}_{-216} (3y^{4/3} - \frac{3}{32}y^{2/3})(4y^{1/3} + \frac{1}{16}y^{-1/3}) [/tex]

from there I tried to multiply out the equation and solve the integral with the bounds, but it isn't giving me the correct answer. I'm not sure what I'm doing wrong. I suspect I have to break the integral up smaller pieces but I am not sure where to break it at.
 
Last edited:
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  • #2
[tex]\int^a_b 2\pi f(y) \sqrt{1+\left(\frac{dx}{dy}\right)^2}[/tex]

you have substituted dx/dy incorrectly into the equation

dx/dy should be inside the root
 
  • #3
Because of the symmetry, you can integrate from 0 to 216 and then multiply by 2. What do you get?
 
  • #4
ok... when I multiply out the integrand I get:

[tex]2\pi [12y^{5/3} - \frac{3}{16}y^2 - \frac{9}{2048}y^{1/3}]^{216}_{0}[/tex]

when I evaluate at 216 & 0, i get 7552892.305... I multiply that by 2 (symmetry) and then multiply that by 2pi, which gives me 94925010.28

edit... nevermind, i made a stupid math mistake. Got it right now, thanks a ton HallsofIvy!
 
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Related to Calculating Surface Area of a Revolution Rotated About the Y-Axis

1. What is the surface area of a revolution?

The surface area of a revolution is the total area of the curved surface formed when a two-dimensional shape is rotated around a fixed axis in three-dimensional space.

2. How is the surface area of a revolution calculated?

The surface area can be calculated by using the formula: S = 2π∫y * √1 + (dy/dx)^2 dx, where y represents the function of the shape being rotated and the integral is taken over the interval of the shape's boundaries.

3. What is the difference between a revolution and a rotation?

A revolution is when a two-dimensional shape is rotated around a fixed axis in three-dimensional space, resulting in a three-dimensional shape. A rotation is when a shape is turned around a fixed point in two-dimensional space, resulting in the same two-dimensional shape in a different orientation.

4. What are some real-life examples of objects with a surface area of a revolution?

Some examples include a cylinder (formed by rotating a rectangle), a sphere (formed by rotating a circle), and a cone (formed by rotating a right triangle). Other examples can include vases, bottles, and even some types of buildings.

5. Why is the surface area of a revolution important?

The surface area of a revolution is important in various fields, such as engineering, architecture, and physics. It helps in calculating the amount of material needed to create a certain shape, and also plays a role in understanding the structural stability and strength of objects.

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