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P2 - Pure maths - Numerical Integration

The region bounded by the y-axis, the lines y=1, y=3 and the curve with equation y=ln(x^2) is rotated completely about the y axis.

Find the volume so generated.

Use the trapezium rule with 5 strips of equal width to find an estimate of the area of the region as given.


I have got 5.68, the answer is 5.70. I am dubious over my method.

Could someone kindly show the correct method with fulls steps. Thanks
Reply 1
Avatar for Gaz031
Gaz031
15
I also get 5.68.
trapezium rule.gif7.2KB
Reply 2
Avatar for john !!
john !!
14
How would you integrate (ln x)^2? My mind has gone blank ...
Reply 3
Avatar for Gaz031
Gaz031
15
mik1w
How would you integrate (ln x)^2? My mind has gone blank ...


You don't. You need to write it in the form x=f(y) as the area is bounded by the curve and the y axis.
Reply 4
Avatar for fabz
fabz
So what IS the volume generated? Do you have the answers? I get e³-e. But it's probably wrong as I am not known for my great integration skills! :biggrin:
Reply 5
Avatar for Gaz031
Gaz031
15
fabz
So what IS the volume generated? Do you have the answers? I get e³-e. But it's probably wrong as I am not known for my great integration skills! :biggrin:

Almost
v.gif1.4KB
Reply 6
Avatar for fabz
fabz
How could I forget good old pi??? Oh well at least I got the rest right!! That's always nice! :smile: Thanks btw!
mik1w
How would you integrate (ln x)^2? My mind has gone blank ...

Int. (lnx)^2 dx. = Int. (lnx)(lnx) dx.
Let u = lnx -> u' = 1/x
Let v' = lnx -> v = xlnx - x

-> Int. (lnx)^2 dx = lnx(xlnx - x) - Int. (xlnx - x)/x dx.
= x(lnx)^2 - xlnx - Int. lnx - 1 dx.
= x(lnx)^2 - xlnx - Int. lnx dx + Int. 1 dx
= x(lnx)^2 - xlnx + x - xlnx + x + k
= x(lnx)^2 - 2xlnx + 2x + k
= x(lnx)^2 + 2x(1 - lnx) + k

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