# Mth101 Questio No2

• Jun 20th 2007, 01:50 AM
TAHIR
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
• Jun 20th 2007, 06:28 AM
Soroban
Hello, TAHIR!

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

Quote:

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?