A2 help please Watch

ADotCross
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Can you help me please

question: the diagram shows the curve y = 6/(3x+1)^(1/2). The shaded region is bound by the curve and the lines
x=2 x=9 and y= 0

ii) the shaded region is rotated completely about the x- axis. Show that the volume of the solid produced can be written in the form of k(ln)2 where the exact value of the constant k is to be determined.
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Mr M
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(Original post by ADotCross)
Can you help me please

question: the diagram shows the curve y = 6/(3x+1)^(1/2). The shaded region is bound by the curve and the lines
x=2 x=9 and y= 0

ii) the shaded region is rotated completely about the x- axis. Show that the volume of the solid produced can be written in the form of k(ln)2 where the exact value of the constant k is to be determined.
Could you show some working or say what you have tried?
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ADotCross
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(Original post by Mr M)
Could you show some working or say what you have tried?
So I used the formula (pi) (intergrate) y^2

so y^2= 36/(3x+1)

then intergration

= (pi)36(1/3) x (1/3x+1)

then simlplify

12(pi)/3x +1

this is is where I got to ( I think I made a mistake in the intergration but dunno where)
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TenOfThem
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(Original post by ADotCross)
So I used the formula (pi) (intergrate) y^2

so y^2= 36/(3x+1)

then intergration

= (pi)36(1/3) x (1/3x+1)

then simlplify

12(pi)/3x +1

this is is where I got to ( I think I made a mistake in the intergration but dunno where)
Indeed - do you know what \displaystyle \int \dfrac{1}{x} dx is
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ADotCross
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(Original post by TenOfThem)
Indeed - do you know what \displaystyle \int \dfrac{1}{x} dx is

Ln(x)

Should it be 12(pi)ln(3x+1) ?
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TenOfThem
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(Original post by ADotCross)
Ln(x)
So, why is the no ln(3x+1) in your integration since you were integrating \dfrac{1}{3x+1}
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TenOfThem
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(Original post by ADotCross)
Ln(x)

Should it be 12(pi)ln(3x+1) ?
yes
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ADotCross
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How do I simplify this ?
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Hasufel
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leaving out the factor of 12Pi for now, what is : F(9)-F(2) ?

(use the law of logs: ln(a^{b})= bln(a)
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TenOfThem
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(Original post by ADotCross)
How do I simplify this ?
Put 9 and 2 in
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ADotCross
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So would it be

12(pi) x (ln28 -ln7) ?
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Hasufel
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yup - now simplify the logs: ln(a)-ln(b) = ln(a/b)

(sorry TenofThem - hijacked your answer somewhat)
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ADotCross
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I suck with logs....

so then that would happen right ?

12(pi)(ln28/7) ?
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ADotCross
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Can I make it into 12(pi) ln4 ?
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davros
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(Original post by ADotCross)
Can I make it into 12(pi) ln4 ?
Yes, and you can write ln4 in terms of ln2
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ADotCross
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So it would become

12(pi)(ln2+ln2)
=
12(pi)ln2 + 12(pi)ln2

do I add or was it times the 12 together

so so it would make it 24(pi)ln2 or is it 144(pi)ln2 ?
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ADotCross
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#17
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Thank you guys so much !!!
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