Tiffany W. answered • 03/24/20
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4e-4x = u
e-4xdx = du/-16
S= 2∏ ∫ (√1+u2 ) (du/-16) [the integral is 0 to 4]
= ∏/16 [4√17 + ln(4√17)]
Tiffany W.
tutor
Yes, but then the surface area of the surface obtained by rotating the curve y=e-4x on (0,infinity ) about x-axis changes as we take the interval from 0 to infinity of 2pi y sqrt(1+(dy/dx)2 dx
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03/24/20
Tiffany W.
tutor
I apologize for not making my work more clear as it's hard to show the math just typing. You may message me and I'd be happy to show you my steps
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03/24/20
Drew M.
Can you please explain to me how to find new limits to integration? Been looking on multiple websites with similar type or problem but I am stuck on 2pi\int(0 to infinity)e^-4xsqrt(1+-4e^-8x)dx-----> e^-4x=u and -4e^-4xdx=du. I'm stuck at this part. How do i find new limits of integration from here?
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03/24/20
Drew M.
Your guidance has been much appreciated at this point. Although do I need to pay or something in order t o message you? I just want t find the new bounds really and I am all good. I just wish I can understand this. I know I just have to find u when x=0 but I get kind of confused here. I'm all out of my free college tutoring sessions so I'm so worried on how I'm going to pass this class due to the online tutoring they let me utilize
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03/24/20
Tiffany W.
tutor
No need! I just messaged you back :)
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03/24/20
Drew M.
Thank you for your help! The answer was correct I just wish I understood the part where you created the new bounds
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03/24/20
Marta B.
Hey Tiffany, could you help me to get the answer for y=e^-x? I did u=e^-2x Since dy/dx square plus one is 1+e^-2x
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03/08/23
Drew M.
isn't dy/dx = -4e^-4x?03/24/20