Calculus - cylindrical shells method?

Question

Use the method of cylindrical shells to find the volume of the solid rotated about the line x = -1 given the conditions: y=x^3 -x^2; y=0; x=0.

Answer 1

is this your homework?

do you like science and math?

Answer 2

this problem is much simpler if you first shift the graph 1 unit to the right so that revolution occurs about the y axis. f(x-h) shifts f h units to the right

y(x)=x^3-x^2 let h=1 y(x-h)=y(x-1)=(x-1)^3-(x-1)^2 use algebra to expand and simplify. so (x-1)^3=x^3-3x^2+3x-1 and
-(x-1)^2= -(x^2-2x+1)= -x^2+2x-1 yielding x^3-4x^2+5x-2=y(x+1)


find y(-1) to get point where y(x) intersects the line x= -1 y(-1)=-2
so x=0 for y(x-1)=-2 and x=0 is intersection of y(x) and y=0 so x=1 for y(x-1)=-2 thus limits of integration are x=0 to 1and new function is Y(x)=x^3-4x^2+5x-2

int (f,x,a,b) denotes integral of f with respect to x from a to b

shell method is int(2*pi*x*f(x),x,a,b)

your problem is int(2*pi*x*(x^3-4x^2+5x-2),x,0,1) distribute x to get int(2*pi*(x^4-4x^3+5x^2-2x),x,0,1)=2*pi* int(x^4-4x^3+5x^2-2x,x,0,1) gives 5.72 units^3 (note ans. was negative since we did not shift graph 2 units up as well so throw out the "-" sign in your final answer because volume is non-negative)

i claim the answer is about 5.72 you'll have to do the final evaluation at limits by hand to get an exact answer