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# Help with C4. :) watch

1. Thanks,
2. (Original post by apo1324)
Hello there,

Could someone please check if my answer is correct for the following question:

The region bounded by the curve y = sqrtx + 4/sqrtx, the x-axis and the lines x = 1, x = 4 is rotated through four right-angles about the x-axis. Find, correct to one decimal place, the volume of the solid formed.

I got an answer of 29.7 (1dp)?? Thank you very much. PS If it is wrong I will post my workings.
Sorry, you're wrong.

If that's then Wolfram says it's 168.6, and I'm inclined to believe it, as I get the same value.
3. When I did it, I got:

y^2 = x + 16/x??

integrating:- x^2 / 2 + 16ln(x)?? I've probably gone wrong bec I'm crap at integrating. Any help?
4. (Original post by apo1324)
When I did it, I got:

y^2 = x + 16/x??

integrating:- x^2 / 2 + 16ln(x)?? I've probably gone wrong bec I'm crap at integrating. Any help?
. use binomial expansion if you must, but is not
5. Cant be arsed to use latex but here we go...

Pi multiplied by the intergral of (rootx + 4/rootx)^2 = what you want.

Multiply out the bracket and you get x + 16/x + 8

Intergrating gives 0.5x^2 + 16lnx + 8x

Applying the limits gives (62.18 - 8.5)pi

=168.64
6. (Original post by ghostwalker)
Sorry, you're wrong.

If that's then Wolfram says it's 168.6, and I'm inclined to believe it, as I get the same value.
this is correct I also got 168.6 (1dp)
7. (Original post by Ollie901)
Cant be arsed to use latex but here we go...

Pi multiplied by the intergral of (rootx + 4/rootx)^2 = what you want.

Multiply out the bracket and you get x + 16/x + 8

Intergrating gives 0.5x^2 + 16lnx + 8x

Applying the limits gives (62.18 - 8.5)pi

=168.64
The bold part... would you not get 'x + 16x^(3/2)' ???
8. (Original post by Mobs25)
The bold part... would you not get 'x + 16x^(3/2)' ???
No you wouldn't. Look again.

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