Area of a surface of revolution

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soleilion

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Find exact area of the surface generated by revolving the curve about the stated axis
1. y=?(x)-1/3 * x^(3/2) , x=[1,3], x-axis

2. 8xy^2= 2y^6+1, y=[1,2], y-axis


I know use which formula, I dont know how to fix them after use formula
please help me
 
soleilion said:
I know use which formula, I dont know how to fix them after use formula
Click to expand...
Please reply showing how far you've gotten in applying the formula. Thank you! :D

Eliz.
 
Is this the equation you are asking about in #1?

\(\displaystyle $x^{3\over 2}\cdot \sqrt{x-{1\over 3}}$\)
 
soleilion said:
can somebody help me right now?
Click to expand...
Sure! Please reply with the formula you're supposed to use, along with a clear listing of all of the steps you've done so far, so we can see where you're having trouble.

Thank you! :D

Eliz.
 
soleilion said:
Find exact area of the surface generated by revolving the curve about the stated axis
1. y=?(x)-1/3 * x^(3/2) , x=[1,3], x-axis

First draw sketch of the function using your graphing calculator.

If you use disk method

volume of your disk = ? * y^2 * dx = ? * [?(x)-1/3 * x^(3/2)] ^2 * dx

Now integrate the above from x=1 to x=3

2. 8xy^2= 2y^6+1, y=[1,2], y-axis

This one first I'll switch x and y

8yx^2 = 2x^6 + 1, x=[1,2], x-axis

We can see x=0 is excluded from consideration. Then

y = 1/4 * x^4 + 1/8 * x^(-2)

Now plot it.

This will give you an idea about the graph for the original expression (which is not a function). What have you been taught about such problems?



I know use which formula, I dont know how to fix them after use formula
please help me
Click to expand...
 
1. 2TT?(x^1/2-1/3*x3/2) ?(1+1/4x^3(1-x)^2) dx =2TT??(x(1-1/3x)^2+1/4x^4(1-!/3x)^2(1-x)^2) dx
x=[1,3]

2. 2TT?(1/4y^4+1/(8y^2))?(1+(y^3-1/(4y)^2)) dy
y=[1,2]

this is what I have right now
please help me
thank you
 
This is formula what i use

s=?2TTy?(1+(y')^2) dx from a to b


s=?2TTx?(1+(x')^2) dy from c to d
 
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