First we need to work out where the two curves intersect one another.
y=x² (upright parabola with vertex at (0,0)) and x=y² (parabola on its side with arms pointing to the right with vertex at (0,0), so, substituting for y in x=y², x=x⁴, x⁴-x=0=x(x³-1); solving for x we get x=0 and x=1.
So the intersection points are (0,0) and (1,1).
(The graph shows the shaded area in the first (positive) quadrant to be rotated around x=-6. This should help you understand the solution. The arbitrary horizontal red line is an example of how the “flywheel” discs are created. It intersects the shaded area at outer point P and inner point Q, showing the radii formed by joining P and Q to x=-6.)
Consider (1,1). The perpendicular from this point onto the line x=-6 forms the radius of the top open surface of the rotated shape. The length of the radius is 1-(-6)=1+6=7 units (see graph). Consider (0,0). The corresponding radius here has a length of 6 units.These are the radii of the highest and lowest points of the shape.
Now consider a point P(x,y) on y=x². The distance from P to x=-6 is x+6. P is going to be rotated around x=-6, which acts like a spindle. So in relation to the spindle the coords of P are (x+6,y). But y=x², so this is (x+6,x²) because the distance to the x-axis from P is not affected by rotating P.
The part of the sideways parabola x=y² that matters in this problem is the part that lies above the x-axis, so we can rewrite x=y² as y=√x. Consider a point Q(x,y), which becomes Q(x,√x). The x value needs to change to x+6 to form a radius of rotation around x=-6. So with respect to x=-6 the coords of Q are x=x+6 and y=√x.
Next we want to consider the volume of a very thin disc when we rotate P and Q around x=-6. The volume is πr²h where r is the radius and h the height or thickness. Since r is horizontal, h must be vertical and we write h=dy and r=x+6. The volume of the disc is π(x+6)²dy. but this disk is completely solid like a flywheel around the spindle. However, we are going to subtract the volume of an inner flywheel from the volume of the outer flywheel.
The volume of the solid of revolution is the sum of the volumes of the individual thin flywheels. The summation in calculus is integral ∫π(x+6)²dy. But we can’t integrate an expression in x with respect to y, so we need to replace x with a function of y. For the outer parabola, y=x², x=√y (the positive half of the upright parabola) and for the inner parabola, x=y².
So we have two integrals: outer is ∫π(√y+6)²dy, inner is ∫π(y²+6)²dy and the limits are [0,1].
The volume of the rotated solid is ∫π(√y+6)²dy-∫π(y²+6)²dy.
That is: π∫(y+12√y+36)dy-π∫(y⁴+12y²+36)dy.
The integrals can be combined:
π∫(y+12√y-y⁴-12y²)dy=π(y²/2+8√y³-y⁵/5-4y³).
Now apply the limits [0,1]: π(½+8-⅕-4)=4.3π=13.5088 approx. cubic units.
