First, sketch the graphs.
y_1=x^2-x
y_2=3-x^2
x-intercept
y_1=0=> x^2-x=0 And we have that {(x=0),(x=1):}
So intercepts are (0,0) and (1,0)
Get the vertex:
y_1=x^2-x=>y_1=(x-1/2)^2-1/4=>y_1-(-1/4)=(x-1/2)^2
So vertex is at (1/2,-1/4)
Repeat previous:
y_2=0=> 3-x^2=0 And we have that {(x=sqrt(3)),(x=-sqrt(3)):}
So intercepts are (sqrt(3),0) and (-sqrt(3),0)
y_2=3-x^2=>y_2-3=-x^2
So vertex is at (0,3)
Result:
How to get the volume? We shall use the disc method!
This method is simply that: "Volume"=piint_a^by^2dx
The idea is simple, however you've got to use it smartly.
And that's what we're going to do.
Lets call our volume V
=>V=V_1-V_2
V_1=piint_a^b(4-y_1)^2dx
V_2=piint_a^b(4-y_2)^2dx
NB: I'm taking (4-y) because y is only the distance from the x-axis to the curve, whereas we want the distance from the line y=4 to the curve!
Now to find a and b, we equate y_1 and y_2 and then solve for x
y_1=y_2=> 2x^2-x+3=0
=>2x^2+2x-3x+3=0
=>(2x-3)(x+1)=0=>{(x=3/2=1.5),(x=-1):}
Since a comes before b, =>a=-1 and b=1.5
=>V_1=piint_(-1)^(1.5)(4-y_1)^2dx=pi int_-1^1.5(4-x^2-x)^2dx= piint_(-1)^(1.5)(x^2+x-4)^2dx
=>piint(-1)^(1.5)(x^4+3x^3-7x^2-8x+16)dx=pi[x^5/5+x^4/2-(7x^3)/3-4x^2+16x]_-1^1.5
V_1=(685pi)/24
Do same for V_2:
V_2=piint_-1^1.5(4-y_2)^2dx= piint_-1^1.5(4-3+x^2)^2dx= piint_(-1)^(1.5)(1+x-4)^2dx
=>piint(-1)^(1.5)(1+2x^2+x^4)dx=pi[x+(2x^3)/3+x^5/5]_-1^1.5
V_1=(685pi)/96
V=V_1-V_2=685/24-685/96=color(blue)((685pi)/32)
