Curved surface FP2

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Bobjim12
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#1
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#1
http://filestore.aqa.org.uk/subjects...W-QP-JUN11.PDF

Question 5.

I'm sitting on S = 2πy∫√(1 +x^2/(x^2+8))

And i'm struggling to get to what they want me to show.

Help pls
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Zacken
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#2
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(Original post by Bobjim12)
http://filestore.aqa.org.uk/subjects...W-QP-JUN11.PDF

Question 5.

I'm sitting on S = 2πy∫√(1 +x^2/(x^2+8))

And i'm struggling to get to what they want me to show.

Help pls
I'm very alarmed by the fact that you have the y outside the integral sign... correct version:

\displaystyle 2\pi \int_0^6 \sqrt{x^2 + 8} \sqrt{\frac{x^2 + 8 + x^2}{x^2 + 8}} \, \mathrm{d}x = 2\pi \int_0^6 \sqrt{2(x^2 + 4)} \, \mathrm{d}x
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Bobjim12
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#3
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(Original post by Zacken)
I'm very alarmed by the fact that you have the y outside the integral sign... correct version:

\displaystyle 2\pi \int_0^6 \sqrt{x^2 + 8} \sqrt{\frac{x^2 + 8 + x^2}{x^2 + 8}} \, \mathrm{d}x = 2\pi \int_0^6 \sqrt{2(x^2 + 4)} \, \mathrm{d}x
i can never tell when we can just ignore the minus root, in this case y has two values............ this makes me sad.
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Zacken
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(Original post by Bobjim12)
i can never tell when we can just ignore the minus root, in this case y has two values............ this makes me sad.
It would have made no difference had you taken the negative root - you'd get a negative area from which (I hope) you'd take the absolute value of to get the positive answer anyway. But at A-Level, you're nearly always going to be taking the positive root without any justification.

In either case - y is inside the integral!
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Bobjim12
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#5
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(Original post by Zacken)
It would have made no difference had you taken the negative root - you'd get a negative area from which (I hope) you'd take the absolute value of to get the positive answer anyway. But at A-Level, you're nearly always going to be taking the positive root without any justification.

In either case - y is inside the integral!
My head hurts.

ok ty
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