What is the surface area when rotating a curve about the x-axis?

In summary, the arc length formula is used to find the length of an arc on a circle and is represented as L = rθ. The central angle for the formula can be found using θ = s/r or by measuring with a protractor. This formula can only be used for circular shapes and the units for the radius and angle should be consistent. There are online calculators and tools available for calculating arc length, but it is important to double check the results for accuracy.
  • #1
sherlockjones
31
0
1 Find the area bounded by the curve [tex] x = t - \frac{1}{t} [/tex], [tex] y = t + \frac{1}{t} [/tex] and the line [tex] y = 2.5 [/tex].

I know that [tex] A = \int_{\alpha}^{\beta} g(t)f'(t) \; dt [/tex]I ended up with [tex] \int_{1}^{2} 2.5-(t+\frac{1}{t})(1+\frac{1}{t^{2}}) [/tex] 2 Find the length of the curve: [tex] x = a(\cos \theta + \theta \sin \theta) [/tex], [tex] y = a(\sin \theta-\theta \cos \theta) [/tex], [tex] 0\leq \theta\leq \pi [/tex]

I obtained [tex] \frac{a\pi^{2}}{2} [/tex]. Does this look correct? I used the arc length formula for parametric equations.Is this correct? 3 Find the surface area obtained by rotating the given curve about the x-axis: [tex] x = 3t-t^{3} [/tex] [tex] y = 3t^{2} [/tex], [tex] 0\leq t\leq 1 [/tex].

So [tex] S = \int_{a}^{b} 2\pi y \sqrt{(\frac{dx}{dt}^{2})+(\frac{dy}{dt}^{2})} \; dt [/tex]

So would I do the following: [tex] \int_{0}^{1} 2\pi(3t^{2})\sqrt{(3-3t^{2})+36t^{2}} \; dt [/tex]?
 
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  • #2
sherlockjones said:
1 Find the area bounded by the curve [tex] x = t - \frac{1}{t} [/tex], [tex] y = t + \frac{1}{t} [/tex] and the line [tex] y = 2.5 [/tex].

I know that [tex] A = \int_{\alpha}^{\beta} g(t)f'(t) \; dt [/tex]


I ended up with [tex] \int_{1}^{2} 2.5-(t+\frac{1}{t})(1+\frac{1}{t^{2}}) [/tex]
Now quite. if x= f(t), y= g(t), then dx= f'(t)dt and ydx= g(t)f'(t)dt but you want (2.5- y)dx. You should have (2.5- t+ 1/t)(1+ 1/t2)dt. Also, recheck your limits of integration. When t= 1, y= 2, not 2.5.


2 Find the length of the curve: [tex] x = a(\cos \theta + \theta \sin \theta) [/tex], [tex] y = a(\sin \theta-\theta \cos \theta) [/tex], [tex] 0\leq \theta\leq \pi [/tex]

I obtained [tex] \frac{a\pi^{2}}{2} [/tex]. Does this look correct? I used the arc length formula for parametric equations.


Is this correct?
Yes, it is. Don't you just love it when things cancel out?


3 Find the surface area obtained by rotating the given curve about the x-axis: [tex] x = 3t-t^{3} [/tex] [tex] y = 3t^{2} [/tex], [tex] 0\leq t\leq 1 [/tex].

So [tex] S = \int_{a}^{b} 2\pi y \sqrt{(\frac{dx}{dt}^{2})+(\frac{dy}{dt}^{2})} \; dt [/tex]

So would I do the following: [tex] \int_{0}^{1} 2\pi(3t^{2})\sqrt{(3-3t^{2})+36t^{2}} \; dt [/tex]?
No. You forgot a square: it should be
[tex] \int_{0}^{1} 2\pi(3t^{2})\sqrt{(3-3t^{2})^2+36t^{2}} \; dt [/tex]
(Once again, that square root simplifies nicely. Your teacher is being nice to you!)
 
  • #3
Instead if the question read as:

Find the surface area generated by rotating the given curve about the y-axis:

[tex] x = 3t^{2}, y = 2t^{3}, 0\leq t \leq 5 [/tex] would it be:

[tex] \int_{0}^{5} 2\pi(3t^{2})\sqrt{72t^{2}} \; dt [/tex]
 
  • #4
No, it wouldn't. You want [itex]\sqrt{(dx/dt)^2+ (dy/dt)^2}[/itex]. Here, dx/dt= 6t and dy/dt= 6t2. You need [itex]\sqrt{36t^2+ 36t^4}= 6t\sqrt{1+ t^2}[/itex].
 

Related to What is the surface area when rotating a curve about the x-axis?

1. What is the arc length formula?

The arc length formula is used to calculate the length of an arc on a circle. It is represented as L = rθ, where L is the arc length, r is the radius of the circle, and θ is the central angle in radians.

2. How do I find the central angle for the arc length formula?

To find the central angle, you can use the formula θ = s/r, where s is the length of the arc and r is the radius of the circle. You can also use a protractor to measure the angle directly from the center of the circle.

3. Can the arc length formula be used for any shape?

No, the arc length formula can only be used for circular shapes. For other shapes, such as ellipses or irregular curves, a different formula would need to be used to calculate the arc length.

4. What units should I use for the radius and angle in the arc length formula?

The units used for the radius and angle in the arc length formula should be consistent. For example, if the radius is measured in meters, the angle should be in radians. If the radius is in inches, the angle should be in degrees.

5. Is there a calculator or tool available to help with calculating arc length?

Yes, there are many online calculators and tools available that can help with calculating arc length. Some graphing calculators also have the capability to calculate arc length. It is important to double check the results from these tools to ensure accuracy.

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