Find the volume of the solid obtained by rotating the region bounded by the given curves about the specified axis.

y=x^2 x=y^2 about the axis x=–5

To find the volume of the solid obtained by rotating the region bounded by the curves y = x^2 and x = y^2 about the axis x = -5, we can use the method of cylindrical shells.

First, let's sketch the region bounded by the curves y = x^2 and x = y^2:

Since x = y^2 is symmetric with respect to the y-axis, we only need to consider the right half of the region. The region bounded by these curves is a symmetrical loop.

Now, let's define a variable for the distance between the axis of rotation (x = -5) and the curve x = y^2. We'll call this distance "r".

To find r, we need to solve the equation x = y^2 for y. Since we only need the right half of the region, we take the positive square root:

y = √x

The distance between x = -5 and the curve x = y^2 is given by:

r = x - (-5) = x + 5

Now, we need to find the height of each cylindrical shell. This is given by the vertical distance between the curves y = x^2 and x = y^2. We can find this by subtracting the equation of the lower curve from the equation of the upper curve:

h = x^2 - y^2

So, the volume of each cylindrical shell is given by:

dV = 2πrh dx

Now, we integrate the volume of each cylindrical shell over the interval where the curves intersect, which is from x = 0 to x = 1:

V = ∫(0 to 1) 2π(x + 5)(x^2 - y^2) dx

To find y^2 in terms of x, we replace y with √x:

V = ∫(0 to 1) 2π(x + 5)(x^2 - √x^2) dx

V = ∫(0 to 1) 2π(x + 5)(x^2 - x) dx

V = ∫(0 to 1) 2π(x^3 + 5x^2 - x^2 - 5x) dx

V = ∫(0 to 1) 2π(x^3 + 4x^2 - 5x) dx

Now, integrate the expression:

V = 2π(1/4)x^4 + (4/3)πx^3 - (5/2)πx^2 | (0 to 1)

V = 2π(1/4)(1)^4 + (4/3)π(1)^3 - (5/2)π(1)^2 - [2π(1/4)(0)^4 + (4/3)π(0)^3 - (5/2)π(0)^2]

V = π/4 + 4π/3 - (5π)/2

V = (π - 10π + 12π)/12

V = 3π/12

V = π/4

Hence, the volume of the solid obtained by rotating the region bounded by y = x^2 and x = y^2 about the axis x = -5 is π/4 cubic units.