19. \( y=x^{3}, y=\sqrt{x} ; \quad \) about the \( x \)-axis

To find the volume of the solid generated by rotating the region bounded by the curves \( y = x^3 \) and \( y = \sqrt{x} \) about the \( x \)-axis, we can use the method of washers. ### Step 1: Find the points of intersection We need to find the points where \( y = x^3 \) and \( y = \sqrt{x} \) intersect. Setting the equations equal to each other: \[ x^3 = \sqrt{x} \] Squaring both sides to eliminate the square root gives: \[ x^6 = x \] Rearranging this, we have: \[ x^6 - x = 0 \] Factoring out \( x \): \[ x(x^5 - 1) = 0 \] This gives us: \[ x = 0 \quad \text{or} \quad x^5 = 1 \implies x = 1 \] Thus, the points of intersection are \( x = 0 \) and \( x = 1 \). ### Step 2: Set up the volume integral The volume \( V \) of the solid of revolution about the \( x \)-axis can be calculated using the washer method. The volume is given by the integral: \[ V = \pi \int_{a}^{b} \left( R(x)^2 - r(x)^2 \right) \, dx \] where \( R(x) \) is the outer radius and \( r(x) \) is the inner radius. In this case: - The outer radius \( R(x) \) is given by \( y = \sqrt{x} \). - The inner radius \( r(x) \) is given by \( y = x^3 \). Thus, we have: \[ R(x) = \sqrt{x}, \quad r(x) = x^3 \] ### Step 3: Calculate the volume The volume integral from \( x = 0 \) to \( x = 1 \) is: \[ V = \pi \int_{0}^{1} \left( (\sqrt{x})^2 - (x^3)^2 \right) \, dx \] This simplifies to: \[ V = \pi \int_{0}^{1} \left( x - x^6 \right) \, dx \] Now, we can compute the integral: \[ V = \pi \left[ \frac{x^2}{2} - \frac{x^7}{7} \right]_{0}^{1} \] Evaluating at the bounds: \[ V = \pi \left( \left( \frac{1^2}{2} - \frac{1^7}{7} \right) - \left( \frac{0^2}{2} - \frac{0^7}{7} \right) \right) \] This simplifies to: \[ V = \pi \left( \frac{1}{2} - \frac{1}{7} \right) \] Finding a common denominator (which is 14): \[ \frac{1}{2} = \frac{7}{14}, \quad \frac{1}{7} = \frac{2}{14} \] Thus: \[ \frac{1}{2} - \frac{1}{7} = \frac{7}{14} - \frac{2}{14} = \frac{5}{14} \] So the volume is: \[ V = \pi \cdot \frac{5}{14} = \frac{5\pi}{14} \] ### Final Answer The volume of the solid generated by rotating the region bounded by \( y = x^3 \) and \( y = \sqrt{x} \) about the \( x \)-axis is: \[ \boxed{\frac{5\pi}{14}} \]