Volume generated by rotating

Question

What is the volume of the solid made by rotating the region bounded by y=x^2+1 and y=-2x+3 about the x - axis. Please show the process using (i) disks/washers, (ii) cylindrical shells

Answer 1

I feel, there is a correction in the integral expression

 

pi* integral (4x^2-12x+9-x^4-2x^2-1)

 

pi *integral(-x^4+2x^2-12x+8)

 

pi(((-x^5)/5) +(2/3)*(x^3)-6*(x^2)+8*x)

 

Limits are same as given by Muhammad

 

So, rest can be solved as usual after applying limits.

 

The formulation of integral and its solution is not that easy using cylindrical shell method.

Answer 2

To solve this, we must first find the two points in which the function intersect. Then we must use the volume integral to find the volume of the solid generated. 

 

x² + 1 = -2x + 3

 

x² + 2x +1 = 3

 

(x+1)² = 3

 

x + 1 = ±√3

 

x = -1 ± √3

 

The linear function is above the parabola for this interval

 

So the volume integral is such:

 

π ∫ A(x) dx =

 

π ∫ (-2x + 3)² - (x² + 1)² dx

 

π ∫ 4x² - 12x + 9 - (x⁴ - 2x² + 1) =

 

π ∫ -x⁴ + 6x² -12x + 8 dx = 

 

π(-x⁵/5 + 2x³ - 6x² + 8x) x=-2.732 to x=.732

 

π[(-(.732)⁵/5 + 2(.732)³ - 6(.732)² + 8(.732)] - π[-(-2.732)⁵/5 + 2(-2.732)³ - 6(-2.732)² + 8(-2.732)]

 

π(-.0420 + .7846 - 3.2154 + 5.8564) - π(30.4420 - 40.7846 - 44.7846 - 21.8564) = 

 

3.3854π + 76.9836π = 80.369π