Find volume of solid obtained by rotating the region bounded by f(x)=3-x^2 & g(x)=1+x^2 about the x-axis?

Question

Find the volume of the solid obtained by rotating the region bounded by the functions f(x)=3-x^2 and g(x)=1+x^2 about the x-axis.
1) How can I find the x-values where the functions intersect?
2) What is the volume of the solid?

Answer 1

1) - this is, of course, fundamental to working out the volume! Fortunately this bit is easy... simply equate f(x) and g(x) and solve for x:
f(x) = g(x)
<=> 3 - x^2 = 1 + x^2
<=> 2x^2 = 2
<=> x = +/- 1.

2) Fortunately both f and g are entirely above the x-axis in this region, with f greater than g; so there's nothing tricky about the volume. It is just
V = π int(-1 to 1) (f(x)^2 - g(x)^2) dx
= π int(-1 to 1) (9 - 6x^2 + x^4 - 1 - 2x^2 - x^4) dx
= π int(-1 to 1) (8 - 8x^2) dx
= 8π [x - x^3 / 3] [-1 to 1]
= 8π (1 - 1/3 - (-1) + (-1/3))
= 8π (4/3)
= 32π/3.

Answer 2

Find the volume of the solid obtained by rotating the region bounded by the functions f(x)=3-x^2 and g(x)=1+x^2 about the x-axis.
1) How can I find the x-values where the functions intersect?

Solve f(x) = g(x)
x = ±1


2) What is the volume of the solid?

V = ∫π[f(x)^2 - g(x)^2] dx [x:-1...1] = 33.51032

Answer 3

So 3-xx=1+xx, hence xx=1, x1=-1, x2=+1 two intersections;
So 3-xx= 0, hence x3=sqrt(3), x4=-sqrt(3); 1+xx>0, thus no x-intercept;
as both functions are even, consider intervals 0<=x<=1 and 1<=x<=sqrt(3), then double; elementary volume dv= (pi*y^2)*dx;
V1= pi* integral{for x=0 until 1} of (1+xx)^2 dx = x +(2/3)x^3 +(1/5)x^5 = {for x=0 until 1} = (1+2/3 +1/5) -0 = 28pi/15;
V2= pi* integral{for x=1 until sqrt3} of (3-xx)^2 dx = 9x –2x^3 +(1/5)x^5 = {for x=1 until sqrt3} = (9sqrt3 –6sqrt3 +(9/5)sqrt3) –(9 –2 +1/5) = pi*1.113844;
V=2(v1+v2) = 2pi*4.094354 =25.726; check me!

Answer 4

the quantity is given by skill of ?? y^2 dx first we desire y^2 y^2 = (4(a million-x^2))^2 = sixteen(x^4 -2x^2 +a million) y^2 = 16x^4 -32x^2 +sixteen for this reason we would desire to do ?? 16x^4 -32x^2 + sixteen dx = ? (3.2x^5 -(32x^3)/3 +16x +c) it incredibly is purely the final expression, in case you had limits you will have an exact quantity and can overlook with regard to the +c consistent on the top too :)