Can someone check my work please?

Question

itz_sid · Answer

The region is bounded by y=ln7x, y=3, y=6; and his revolved around the y-axis.
$$\pi \int\limits_{0}^{6} (\ln7x)^2 dx$$
Then we do Integration by Parts...
$$u = (\ln7x)^2 \rightarrow du= \frac{ 2\ln(7x) }{ x }$$
$$dv = dx \rightarrow v = x$$
So we get...
$$xln^27x-2\int\limits_{0}^{6}\ln7xdx$$
Is that correct so far?

3mar · Answer

by the way, I was solving this problem when you posted it

3mar · Answer

this is correct and what is after that?

mathmale · Answer

In your integral you have not written "x" in front of "2(ln 7x)."  Think about this.  Understand where that "x" comes from?    How would you simplify ln (7x)?

irishboy123 · Answer

that's the revolution about the x-axis of the wrong area

agent0smith · Answer

You should start with a diagram...

itz_sid · Answer

@mathmale Um... Idunno. :/
@IrishBoy123 Oh it should be from 3 to 6 huh?
@agent0smith A diagram? I wasn't taught how to do it that way. Could you explain?

itz_sid · Answer

Oh wait and i should change my equations and solve for x to plug in

zepdrix · Answer

`The region is bounded by y=ln7x, y=3, y=6`

This does not create a region.
Are you sure you didn't miss another boundary?

zepdrix · Answer

Is the y-axis the fourth boundary?

itz_sid · Answer

and x=0

zepdrix · Answer

k

itz_sid · Answer

Yea

agent0smith · Answer

A graph...? Should be your first thing.

itz_sid · Answer

Oh i graphed it in my notebook.|dw:1474750632886:dw|