To find the volume $V$ of the solid formed by rotating the region bounded by the curves $y = x^2$ and $x = y^2$ around the line $y = 1$, we will use the method of cylindrical shells.

Step 1: Set up the region

• The curve $y = x^2$ is a parabola opening upwards.
• The curve $x = y^2$ is a parabola opening sideways.

We need to identify the points of intersection of these two curves. Set $y = x^2$ and $x = y^2$, which gives: $x = (x^2)^2 \quad \Rightarrow \quad x = x^4.$ Solving for $x$, we have: $x(x^3 - 1) = 0 \quad \Rightarrow \quad x = 0 \quad \text{or} \quad x = 1.$ So, the curves intersect at $x = 0$ and $x = 1$.

Step 2: Set up the integral

Using the method of cylindrical shells, the formula for the volume is: $V = 2\pi \int_{a}^{b} ( \text{radius} ) ( \text{height} ) \, dx.$

• The radius of a typical shell is the vertical distance from the curve to the line $y = 1$, which is $1 - y$.
• The height of the shell is the horizontal distance between the two curves, which is $x = y^2$ and $y = x^2$.

We rewrite the problem in terms of $y$, since the rotation is around the line $y = 1$.

Step 3: Boundaries and limits of integration

• For $y = 0$ to $y = 1$, the height of the region is given by the curve $x = y^2$ and $y = x^2$.

Thus, we integrate between $y = 0$ and $y = 1$.

Step 4: Volume calculation

The integral becomes: $V = 2\pi \int_{0}^{1} (1 - y) \left( y^{1/2} - y^2 \right) \, dy.$ Now, we will solve this integral.

Step 5: Solve the integral

Expanding the expression: $V = 2\pi \int_{0}^{1} \left[ (1 - y)(y^{1/2} - y^2) \right] \, dy$ $V = 2\pi \int_{0}^{1} \left[ (y^{1/2} - y^2) - y(y^{1/2} - y^2) \right] \, dy$ $V = 2\pi \int_{0}^{1} \left[ y^{1/2} - y^2 - y^{3/2} + y^3 \right] \, dy.$ Now, integrate each term separately: $V = 2\pi \left[ \frac{2}{3}y^{3/2} - \frac{y^3}{3} - \frac{2}{5}y^{5/2} + \frac{y^4}{4} \right]_{0}^{1}.$ Evaluating at $y = 1$ and $y = 0$: $V = 2\pi \left[ \frac{2}{3} - \frac{1}{3} - \frac{2}{5} + \frac{1}{4} \right].$ Now simplify: [ V = 2\pi \left[ \frac{2}{3} - \frac{1}{3} - \frac{2}{5} + \frac{1}{4} \right] = 2\pi \left[ \frac{1}{3} - \frac{2}{5} + \frac{1}{4} \right]. ] Find a common denominator (60): [ V = 2\pi \left[ \frac{20}{60} - \frac{24}{60} + \frac{15}{60} \right] = 2\pi \left[ \frac{11}{60} \right]. ] So, the volume is: $V = \frac{22\pi}{60} = \frac{11\pi}{30}.$

Final Answer:

The volume of the solid is $V = \frac{11\pi}{30}$.

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Would you like more details or clarifications on any of these steps?

Here are 5 related questions you might want to explore:

1. How do you apply the disk/washer method to solve similar problems?
2. Can this volume be computed using polar coordinates?
3. What is the difference between rotating around $y = 1$ and $y = 0$?
4. How would the volume change if we rotated around a vertical line like $x = 1$?
5. How do you find the surface area of the solid formed by this rotation?

Tip: When using the method of cylindrical shells, focus on how the radius and height change with respect to the axis of rotation to set up the correct integrals.