I need help integrating S = int[a,b] 2 pi x sqrt{ 1 + [f '(x)]^2 } dx

[Violet13]

This is what Im trying to integrate but I dont know how to:

\(\displaystyle \displaystyle \int_0^{179.8}\, \)\(\displaystyle 2\, \pi\, \left(-0.00141x^2\, +\, 0.12320x\, +\, 25.24546 \right)\,\)\(\displaystyle \left( \sqrt{\strut 1\, +\, \left(-0.00282x\, +\, 0.12320 \right)^2\,}\right) \, dx\)

I don't know if you guys can read that so I can break it down for you it is basically the formula for surface area of revultion about the x-axis with f(x) and f'(x) plugged in.

\(\displaystyle S\, =\, \)\(\displaystyle \displaystyle \int_a^b\, \)\(\displaystyle 2\, \pi\, x\, \sqrt{\strut 1\, +\, \left[f'(x)\right]^2\,}\, dx\)

this is the formula for surface area of revolution about the x-axis.

For my situation f(x) is -0.00141x^2+0.1230x+25.24546
a= 0, and b=179.8

If anyone can figure out or tell me how to integrate it step by step it will be super helpful and you will save me from failing a calculus class.

Thanks in advance

 

[Deleted member 4993]

This is what Im trying to integrate but I dont know how to:

\(\displaystyle \displaystyle \int_0^{179.8}\, \)\(\displaystyle 2\, \pi\, \left(-0.00141x^2\, +\, 0.12320x\, +\, 25.24546 \right)\,\)\(\displaystyle \left( \sqrt{\strut 1\, +\, \left(-0.00282x\, +\, 0.12320 \right)^2\,}\right) \, dx\)

I don't know if you guys can read that so I can break it down for you it is basically the formula for surface area of revultion about the x-axis with f(x) and f'(x) plugged in.

\(\displaystyle S\, =\, \)\(\displaystyle \displaystyle \int_a^b\, \)\(\displaystyle 2\, \pi\, x\, \sqrt{\strut 1\, +\, \left[f'(x)\right]^2\,}\, dx\)

this is the formula for surface area of revolution about the x-axis.

For my situation f(x) is -0.00141x^2+0.1230x+25.24546
a= 0, and b=179.8

If anyone can figure out or tell me how to integrate it step by step it will be super helpful and you will save me from failing a calculus class.

Thanks in advance:smile:

f(x) = -0.00141x^2+0.1230x+25.24546

f'(x) = -0.00282 * x + 0.123

[f'(x)]² = ?

and continue...
What are your thoughts?

Please share your work with us ...even if you know it is wrong

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[Violet13]

Integration Help Please

This is the formula for surface area of revolution over the x -axis for my case f(x) is -0.00141x^2+0.12320x+25.24546 , a=0, b=179.8 and I need help algebraically and manually doing the integral and finding the surface area step by step. If I plug it in my calculator I get 24139.1142 as the answer but I need to be able to do it algebraically step my by step.When I plug f(x) and f'(x) in and try to antiderive I get stuck and I dont know how to do it.

This is it written down: \int _0^{179.8}2\pi \left(-0.00141x^2+0.12320x+25.24546\right)\left(\sqrt{1+\left(-0.00282x+0.12320\right)^2}\right)\:dx

If any of you geniuses out there can help me out I would seriously appreciate it. Please help a child in need.

 

[Violet13]

Sorry I am new to this forum and dont really understand how it works so excuse my ignorance. I can derive f(x) myself and then am able to plug it in the SA formula and then move the constant out of the integral but from there I dont know how to antiderive because of the complex nature. I know that I cannot split the integral in two because the parts are being multiplied which is why I get overwhelmed and lost

Thanks for reply so soon!

 

[Deleted member 4993]

f(x) = -0.00141x^2+0.1230x+25.24546

f'(x) = -0.00282 * x + 0.123

-0.00282 * x + 0.123 = tan(Θ)

1 + [f'(x)]² = sec²(Θ)

-0.00282 dx = sec^2(Θ) dΘ

Now continue....

 

[Violet13]

Sorry I may come off as stupid but I have no idea how to continue that. Could maybe please continue it for me? I can maybe figure out how to do it after seeing all the work.