Surface Area of a Solid of Revolution

In summary, the author's attempt to solve a homework equation was unsuccessful because they did not understand the hint and did not include the necessary information in their attempt.
  • #1
Oblakastouf
18
0

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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  • #2
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.
 
  • #3
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.
 
  • #4
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..
 
  • #5
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...
 
  • #6
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.
 
  • #7
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...
 
  • #8
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.
 
  • #9
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.
 

What is the "Surface Area of a Solid of Revolution"?

The surface area of a solid of revolution is the total area of the curved surface formed when a two-dimensional shape is rotated around an axis to create a three-dimensional object.

How is the surface area of a solid of revolution calculated?

The surface area of a solid of revolution can be calculated using the formula 2π∫a^b y√(1+(dy/dx)^2) dx, where a and b are the limits of integration and y is the equation of the curve being rotated.

What types of shapes can be used to create a solid of revolution?

Any two-dimensional shape can be used to create a solid of revolution, as long as it is rotated around an axis. Common examples include circles, ellipses, and parabolas.

Why is the surface area of a solid of revolution important in mathematics?

The surface area of a solid of revolution is important in mathematics because it allows us to calculate the surface area of complex three-dimensional objects, which has practical applications in fields such as engineering and physics.

How is the surface area of a solid of revolution different from the surface area of a regular solid?

The surface area of a solid of revolution is calculated differently from the surface area of a regular solid because it is formed by rotating a two-dimensional shape, rather than having flat sides. This results in a curved surface, which requires a different formula for calculation.

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