Surface Area of a Solid of Revolution

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Homework Help Overview

The discussion revolves around calculating the surface area of a solid of revolution using a specific integral formula. Participants express confusion regarding the provided hint and the complexity of the algebra involved in setting up the integral.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants question the utility of the hint provided, with some suggesting it simplifies the process of finding the derivative. Others express frustration with the algebraic manipulations required and the lack of foundational knowledge in algebra and trigonometry.

Discussion Status

There is an ongoing exploration of different methods to evaluate the integral, with some participants offering guidance on algebraic techniques. The discussion reflects a mix of attempts to understand the problem and varying levels of confidence in handling the mathematical concepts involved.

Contextual Notes

Some participants note gaps in their prior education regarding algebra and trigonometry, which they feel hinder their ability to tackle the problem effectively. There is mention of competition with peers who may have a stronger background in these areas.

Oblakastouf
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Homework Statement



2qlet89.jpg


Homework Equations



integral [a, b] 2*pi*x*sqrt(1 + (dy/dx)^2)

The Attempt at a Solution



Well I know how to do surface area questions... But that the @#$@ is with this random equation? How would I even start to evaluate it... Like honestly... I don't even understand the worthless hint they give.
 
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Did you not observe that the hint gives you a formula for the derivative of your curve, so you don't have to compute it yourself?

Just try writing things out and you should find that the algebra reduces in a clever way.
 
How do you know the hint is worthless if you haven't actually set up the integral that represents the surface area? Or if you have, you didn't include it in your attempt at a solution.
 
I can never see the algebra that they try to make "Clever".
It annoys the hell out of me that they need to make it "Clever" instead of letting us do it the sec^3(x) method... That's easy as hell in comparison to this complex algebra that I never learned because my high school teachers couldn't teach..
 
Here's the integral:

2*pi*y*sqrt(1+(sin(5x)*sqrt(3-cos(5x)^2))^2)

I just don't see an EASY method to evaluate it... All of the method's I've tried end up in just a more and more complicated integral...
 
The y at the beginning should be x.

Expand all the stuff inside the radical, change cos^2(5x) to (1 - sin^2(5x)), and you should get something that is a perfect square.
 
Yeah see, we were never taught what the heck perfect squares are in high school... Basically we were told to use the quadratic equation for everything. We never even learned cubes...
 
I'd advise you to buy or borrow a book or two on algebra and maybe trig to help you learn what you should have been taught in HS. I'm sure they have something on amazon.com or abebooks.com.

Most likely the other people in your class now have these skills, so you're in competition with folks for whom a lot of this stuff is easy. A big part of calculus is being able to turn one expression into another using algebra and trig. If you're weak on those areas, it makes it that much harder to follow what's going on.
 
integral xsin^2(5x) + integral x

Integral x/2 + Integral xcos(10x)/2 + integral x

right?
 
  • #10
Yeah, that looks about right. Don't forget that multiplier of 2pi and that you're working with a definite integral.
 
  • #11
Gawd... That was a f--king ordeal... Thanks though lol.
 

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