volumes of solids of rev.

[mano91]

exercise 8D from essentials book.
question 29.

y=loge(x)       between  x=2  x=3 rotated about the x axis.
find the volume correct to 3 dec. places.

can this be done without a calculator?
ive tried looking at the inverse function and breaking it up into cylinders. but there is this tiny area i dont know how to do.

[TrueTears]

Here's a good trick [Kind of stepping into the territory of integration by parts, check it out in wiki, good to know]









For this question, you don't need the +c

[QuantumJG]

exercise 8D from essentials book.
question 29.

y=loge(x)       between  x=2  x=3 rotated about the x axis.
find the volume correct to 3 dec. places.

can this be done without a calculator?
ive tried looking at the inverse function and breaking it up into cylinders. but there is this tiny area i dont know how to do.

V = pi*int[(ln(x))^2]dx

this can be found without a calculator! But you need a little tertiary calculus.

V = pi*[x*(lnx)^2 - 2int(lnx)dx]

= pi*[x*(lnx)^2 - 2x(lnx - 1)]

V (2 -> 3) = pi*[3(ln3)^2 - 6ln3 + 6 + 4ln2 - 4 - 2(ln2)^2]

= pi*[3ln3(ln3 - 2) - 2ln2(ln2 - 2) + 2]

= 2.642 units cubed (correct to 3 decimal places)

[dino]

isn't it easier to get the inverse, hence an exponential?


area = pi int(x)^2 dy?

or maybe I'm simplifiying things?

[dcc]

The simplest you'll get it to in VCE is , which is still impossible.  Use a calculator .

It's not only easier, but you are less likely to make a mistake / confuse examiner / make a mistake.

[TrueTears]

isn't it easier to get the inverse, hence an exponential?


area = pi int(x)^2 dy?

or maybe I'm simplifiying things?

You can but why would you when you can just integrate it.