1. ## Mth101 Questio No2

Find the volume of the solid that results when the region enclosed by the given curves is revolved about the x-axis.
y=x^2+1, y=x+3

2. Hello, TAHIR!

I assume you are new to this type of problem.
. . I'll give you a leisurely walk-through.

Find the volume of the solid that results when the region enclosed by the given curves
is revolved about the x-axis: . $y\:=\:x^2+1,\;\; y\;=\;x+3$
First, make a sketch . . .

The first is an up-opening parabola with y-intercept 1.
The second is an up-sloping line with y-intercept 3.
Code:
                  |             /
*           |           *
|         /:.
|       /:::.
*        |     /::::*.
|   /:::::  .
| /::::::   .
*     3*::::::*    .
/:|:::::      .
*:::|:::*       .
/   ::|::         .
/   .   *1          .
/     .   |           .
- / - - - + - + - - - - - + -
/        -1   |           2
|
For their intersections, equate the functions.
. . $x^2 + 1 \:=\:x + 3\quad\Rightarrow\quad x^2 - x - 2 \:=\:0\quad\Rightarrow\quad(x + 1)(x - 2) \:=\:0$

Hence, they intersect at: . $x\:=\:-1,\:2$

The "upper" function is: . $y_2 \:=\:x + 3$
The "lower" function is: . $y_1\:=\:x^2 + 1$

The "washer" formula is: . $V \;=\;\pi\int^b_a\left(y_2^2 - y_1^2\right)\,dx$

Hence, we have: . $V \;=\;\pi\int^2_{\text{-}1}\left[(x+3)^2 - (x^2 + 1)^2\right]\,dx$

Can you finish it now?