Integration Question

[fishbiscuit]

For our next assessment task our teacher gave us a "question bank" with 50 questions to do at home, and I'm just confused about one of the questions. I don't have the answers for these questions (because we're going to be tested on some of the questions in class):

i) Sketch the curve y = ln(1+x) from x=0 to x=6

This was all good

ii) The region bounded by this portion of the curve and the y-axis is rotated 360 degrees about the y-axis. Find the volume of the solid of revolution formed. Leave your answer in exact form.

My problem is - what are the boundaries if the question is asking for the volume about the y-axis? I firstly thought maybe if I subbed in x=6 into the original equation to get y = ln(7) so I have 0 and ln(7) as my boundaries... but then I thought that doing that doesn't really makes sense because I'm cutting off the region at ln(7)... so then I thought maybe the volume is infinity? I can't be too sure though.

I'd just like some clarification in case I missed something. Thank you!

[jamonwindeyer]

Hey there!! You were correct to use \(\ln{7}\) as your upper bound. It says by this portion of the curve, so it wants you to stop at the y-value corresponding to \(x=6\)

[fishbiscuit]

Hey there!! You were correct to use \(\ln{7}\) as your upper bound. It says by this portion of the curve, so it wants you to stop at the y-value corresponding to \(x=6\)

I'm glad I was on the right track, but can you please explain  the wording of the question and the "bounded by this portion of the curve" because I'm a bit confused about it.
Sorry
Thank you for your help

[RuiAce]

It just means the region bounded by the curve \( y = \ln (x+1)\) and the \( y\)-axis. As opposed to the region below it, which is still bounded by \( y = \ln (x+1) \), but also by the \(x\)-axis instead.

[jamonwindeyer]

I'm glad I was on the right track, but can you please explain  the wording of the question and the "bounded by this portion of the curve" because I'm a bit confused about it.
Sorry
Thank you for your help

Sure! So like, in Part (a), you drew a curve for a specific domain (0 to 6). You drew a portion of the curve. They then want you to take that portion and spin it around the y-axis. So you aren't spinning around for infinitely large values of y, it only goes up to ln(7). I agree it isn't the clearest of questions, but that's what they mean. It's like the rest of the curve doesn't even exist, so the endpoint coordinate of (6,ln7) becomes your cut off point for any areas (or in this case, volumes)

[fishbiscuit]

Oooooooooooh.
So the graph basically doesn't exist after x=6, that makes sense. Thank you very much!!!