Question

The solid lies between planes perpendicular to the x axis at x=-1 and x=1.the cross-section perpendicular to the x axis are circular disks whose diameter run from parabola y=x2 to the parabola y=2-x2

Answer 1

Answer:

352/105

Step-by-step explanation:

Diameter of the circular disk is

(2−x2)−x2

so that the radius of the disk is the diameter divide by 2

(2−x2−x2)/2.

Integrating V at 1 and -1, we get

V =Z 1−1A dx =Z 1−1π(1 − x2)2

dx = π(x −2/3x3 +1/5x5)1−1

=16/15π

= 352/105

Answer 2

Volume of the solid will be $\frac{16}{15} \pi \ m^3$.

Step 1: Find the area of cross section.

Given - Cross sections are circle whose diameter run from $y=x^2$ and $y=2-x^2$.

$d= 2-x^2-x^2\\=2-2x^2$

$r=\frac{1}{2} d$

$= \frac{1}{2} (2-2x^2)$

$=1-x^2$

Area of cross section = $\pi r^2$

                                   = $\pi (1-x^2)^2$

                                   = $\pi (1-2x^2+x^4)$

Step 2: Find the volume of the solid.

Limitation for integration will be from $x=-1$ to $x=1$.

$Volume= \int\limits {Area} \, dx$

$= \int\limit^1 _-_1 {\pi (1-2x^2+x^4)} \, dx$

$=\pi (x-\frac{2}{3} x^3+\frac{1}{5} x^5)|^1_-_1$

$= \pi [(1-\frac{2}{3} +\frac{1}{5} )-(-1+\frac{2}{3} -\frac{1}{5} )]$

$= \pi (1-\frac{2}{3} +\frac{1}{5} +1-\frac{2}{3} +\frac{1}{5} )$

$= \pi (2-\frac{4}{3} +\frac{2}{5} )$

$=\pi (\frac{30-20+6}{15} )$

$=\frac{16}{15} \pi \ m^3$