- #1
Lo.Lee.Ta.
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- 0
"Find the area of the surface of the curve obtained by rotating the.."
1. Find the area of the surface obtained by rotating the curve y= 1+5x^2
from x=0 to x=5 about the y-axis.
2. I thought to find surface area, we would need to use this formula:
SA= ∫2[itex]\pi[/itex](f(x))√(1 + (f'(x))2)]dx
3. So far I have:
∫0 to 5 of 2[itex]\pi[/itex][(1 + 5x2)(√(1+100x2))]dx
I first tried to do a u substitution for what's in the square root (1 + 100x2), but then du only equals
200x... And I'm not able to replace the 5x^2...
So it looks like u substitution will not work.
So then I thought I might be able to use the product rule.
So I did:
SA= ∫2[itex]\pi[/itex][(1+ 5x^2)(1/2(100x^2)-1/2)*200x + ((1+100x2)1/2)(10x))]dx |0 to 5
But when I plug in 5 and 0 from here, I get a large number.
The number I get is 7521[itex]\pi[/itex].
But when I put this integral into Wolfram Alpha, it says the answer is an even larger number: 49,894!
I am confused as to what I am doing wrong!
Please help!
Thank you so much! :)
1. Find the area of the surface obtained by rotating the curve y= 1+5x^2
from x=0 to x=5 about the y-axis.
2. I thought to find surface area, we would need to use this formula:
SA= ∫2[itex]\pi[/itex](f(x))√(1 + (f'(x))2)]dx
3. So far I have:
∫0 to 5 of 2[itex]\pi[/itex][(1 + 5x2)(√(1+100x2))]dx
I first tried to do a u substitution for what's in the square root (1 + 100x2), but then du only equals
200x... And I'm not able to replace the 5x^2...
So it looks like u substitution will not work.
So then I thought I might be able to use the product rule.
So I did:
SA= ∫2[itex]\pi[/itex][(1+ 5x^2)(1/2(100x^2)-1/2)*200x + ((1+100x2)1/2)(10x))]dx |0 to 5
But when I plug in 5 and 0 from here, I get a large number.
The number I get is 7521[itex]\pi[/itex].
But when I put this integral into Wolfram Alpha, it says the answer is an even larger number: 49,894!
I am confused as to what I am doing wrong!
Please help!
Thank you so much! :)