The Student Room Group

Curved surface FP2

http://filestore.aqa.org.uk/subjects/AQA-MFP2-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
Reply 1
Original post by Bobjim12
http://filestore.aqa.org.uk/subjects/AQA-MFP2-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 yy outside the integral sign... correct version:

2π06x2+8x2+8+x2x2+8dx=2π062(x2+4)dx\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
Reply 2
Original post by Zacken
I'm very alarmed by the fact that you have the yy outside the integral sign... correct version:

2π06x2+8x2+8+x2x2+8dx=2π062(x2+4)dx\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.
Reply 3
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!
Reply 4
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 :smile:

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