Finding the surface area of a 3 dimensional light bulb!

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May 19th 2009, 04:12 AM
markxchan

Finding the surface area of a 3 dimensional light bulb!

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Find the surface area of a 3-dimensional light bulb formed using the following equation:

$y=\frac{1}{3}x^{1/2}-x^{3/2}$ and $0\leq x\leq \frac{1}{3}$
May 19th 2009, 04:48 AM
stapel

Quote:

Originally Posted by markxchan

http://img90.imageshack.us/img90/2697/document.jpg

Find the surface area of a 3-dimensional light bulb formed using the following equation:

$y=\frac{1}{3}x^{1/2}-x^{3/2}$ and $0\leq x\leq \frac{1}{3}$

The surface area will be the sum of all of the circumferences formed by rotating the given curve about the x-axis.

The formula for the circumference C of a circle with radius y is $2\pi y$ . So I think you need to find the integral:

$2\pi\,\int_0^{\frac{1}{3}}\, \left(\frac{1}{3}x^{\frac{1}{2}}\, -\, x^{\frac{3}{2}}\right)\, dx$

(Wink)
May 19th 2009, 04:56 AM
markxchan

Thanks. I would really appreciate it if someone can give me a complete solution too, so that I can check whether I'm doing things right.
May 19th 2009, 04:57 AM
stapel

Quote:

Originally Posted by markxchan

I would really appreciate it if...I can check whether I'm doing things right.

Please show what you've done, and we'll be glad to check it over!

Thank you! :D
May 19th 2009, 11:52 AM
Kaitosan

Stapel, what are you talking about? Surface area is calculated by finding the circumference times arc length (either in terms of dx or dy) within a definite integral.

Based on what the problem looks like, you should use dy. So you need to find the maximum height of y and take 0 and that maximum height as the limits of integration.

Actually, I think you also have to multiply your result by 2 because the dy limits of integration covers only the first half of the light bulb.
May 20th 2009, 05:00 AM
markxchan

kaitosan, can you give the formula of what you are saying so that i can better understand?
May 20th 2009, 05:28 AM
Kaitosan

Surface of revolution - Wikipedia, the free encyclopedia

Check out the first three formulas listed.

Basically, if the equation is parametric equation, use the arc length of parametric equation times 2pi and expressions in terms of x or y depending on the orientation of the radius.

If the equation is in terms of y = f(x), then use the arc length formula for standard rectangular equation times the circumference (either an expression of y or x depending on the orientation of the radius).

Now, here's the thing. Based on your equation, it'll be much easier to use dy because it looks very difficult to isolate x. So, somehow you'll have to manage on using dy for rotation about the x-axis, which is unusual.

Now, I'm not 100% sure on this issue, but I'm pretty sure you take the limits of integration of when the circumference is increasing (from 0 to the maximum value of y). Then either multiply the result by 2 since the expression is symmetric or make another integral with limits of integration from the max value to zero (but make sure to absolute value the whole thing in case...).

There are many smarter people here than me. I'd appreciate if someone confirm what I'm saying.