FP3 surface area of rotation

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#1
paper here:

https://pmt.physicsandmathstutor.com...%20Edexcel.pdf

mark scheme here:

https://pmt.physicsandmathstutor.com...%20Edexcel.pdf

Question 7: It asks for the surface area of a half revolution -- pi radians, yet the working uses 2(pi)r corresponding to what I'm imagining is the full elementary circumference. What am I missing?

Thanks.
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1 year ago
#2
(Original post by jameshyland29)
paper here:

https://pmt.physicsandmathstutor.com...%20Edexcel.pdf

mark scheme here:

https://pmt.physicsandmathstutor.com...%20Edexcel.pdf

Question 7: It asks for the surface area of a half revolution -- pi radians, yet the working uses 2(pi)r corresponding to what I'm imagining is the full elementary circumference. What am I missing?

Thanks.
Circle has an upper semicircle and a lower semicircle.

Rotating through pi radians generates a whole sphere.

Surface area is either area generated by both halves rotated through pi radians, or by one semicircle rotated through 2pi radians.
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#3
Ah of course!! Thanks.
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#4
Now I'm wondering, why does y = (f(x))^0.5 only exist for y > 0 when the square root of f(x) could be negative? I remember hearing about some "convention" of taking the positive square root but I can't see why it's necessary.

Thanks.
0
1 year ago
#5
(Original post by jameshyland29)
Now I'm wondering, why does y = (f(x))^0.5 only exist for y > 0 when the square root of f(x) could be negative? I remember hearing about some "convention" of taking the positive square root but I can't see why it's necessary.

Thanks.
The "square root" is a function - single valued - and as such returns the positive square root.

Which is why when you want both the positive and negative, as in the quadratic formula you have to say +/- sqrt(b^2-4ac).
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#6
Got it. Thanks.
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