Question

The region bounded by y = 1/x, the x-axis and the lines x = 1, x= a (a>1) is rotated about the X-axis. Find volume (V) and hence lim a->infinity V

{pi (1-1/a):pi}

Answer 1

\[V = \int\limits_{1}^{a}\pi \left (\frac 1x\right )^2dx\]

Answer 2

where do we got from there

Answer 3

First integrate the above and find V as a function of 'a'.

Answer 4

wait i integrated the above and got pi [1/x]a,1

Answer 5

how do i get rid of the 'x'

Answer 6

You evaluate 1/x at x = 1 and subtract 1/x at x = a
V = pi * (1 - 1/a) which is answer to the first part.
For the second part, evaluate the limit of V as a->infinity.
a->infinity, 1/a -> 0
So V when a->infinity = pi which is the answer to the second part.

Answer 7

{pi (1-1/a):pi} what is this part

Answer 8

\[V = \pi \left [ \frac 1x \right ]_a^1 = \pi \left [1 - \frac 1a \right ]\]

Answer 9

why is there an extra pi

Answer 10

a ratio?

Answer 11

There are two questions in this problem.

The above answer is to the question: "Find volume (V)"

The second part of the question is: "and hence lim a->infinity V"
So we need to find: \[\lim_{a \rightarrow \infty }\pi \left [ 1 - \frac 1a \right ] = \pi\]

Answer 12

so what working out would we do to find out? sry im rlly noob

Answer 13

We have already done the whole problem!

First question: Find V. \[ V = \pi \left [ \frac 1x \right ]_a^1 = \pi \left [1 - \frac 1a \right ]\]The second question is: Find V as a->infinity:
\[\lim_{a \rightarrow \infty } V= \lim_{a \rightarrow \infty }\pi \left [ 1 - \frac 1a \right ] = \pi\]

Those are the two answers you have. They have just separated the answers using a colon.