Question

a. Find the volume V(S) of the solid S by revolving the region R bounded by x 2 + y − 4 = 0 and x − y + 2 = 0 about y = 0. (6 pts.) b. Set-up the integral that represents V(S) when the region R in (a.) is revolved about x = −2. (4 pts.)

Answer 1

a)

Points of interception:

$x^2+y-4=0\\x-y+2=0$

$x^2+x+2-4=0$

$x=-2$ , $x=1$ .

V(S) can be obtained if we subtract volume

bounded by $x-y+2=0$ , $y=0$ , $-2\le x\le1$ and revolved about $y=0$

from the volume

bounded by $x^2+y-4=0$ , $y=0$ , $-2\le x\le1$ and revolved about $y=0$

$V(S)=\pi\int_{-2}^1((-x^2+4)^2-(x+2)^2)dx$ $=\pi\int_{-2}^1(x^4-8x^2+16-x^2-4x-4)dx$

$=\pi(x^5/5-9x^3/3-4x^2/2+12x)|_{-2}^1$ $=\pi(1/5-3-2+12+32/5-3\cdot8$$+2\cdot4+12\cdot2) =108\pi/5$

b)

Let’s move to new coordinates:

$t=x+2$

u=y

Now region R is bounded by

$(t-2)^2+u-4=0$ and $t-u=0$

V(S) is obtained by revolving R about $t=0$

Look at the picture. Region R is divided into R1 and R2.

Volume V1 is obtained by revolving R1 about $t=0$ can be calculated by subtracting volume

bounded by $t=2-\sqrt{4-u}$ , $t=0$ , $0\le u\le3$ and revolved about $t=0$

from the volume

bounded by $t=u$ , $t=0$ , $0\le u\le3$ and revolved about $t=0$.

$V1=\int_0^3\pi(u^2-(2-\sqrt{4-u})^2)du$

Volume V2 is obtained by revolving R2 about $t=0$ can be calculated by subtracting volume bounded by $t=2-\sqrt{4-u}$ , $t=0$ , $3\le u\le4$ and revolved about $t=0$

from the volume

bounded by $t=2+\sqrt{4-u}$ , $t=0$ , $3\le u\le4$ and revolved about $t=0$.

$V2=\int_3^4\pi((2+\sqrt{4-u})^2-(2-\sqrt{4-u})^2)du$

$V(S)=V1+V2=$$\int_0^3\pi(u^2-(2-\sqrt{4-u})^2)du+$$\int_3^4\pi((2+\sqrt{4-u})^2-(2-\sqrt{4-u})^2)du$