Let \( R \) be the region bounded by the curves \( y=x \) and \( y=x^{\frac{1}{3}} \). Let \( S \) be the solid generated when \( R \) is revolved about the \( x \)-axis in the first quadrant. Find the volume of \( S \) by both the disc/washer and shell methods. Check that your results agree.

To find the volume of the solid \( S \) generated by revolving the region \( R \) bounded by the curves \( y = x \) and \( y = x^{\frac{1}{3}} \) about the \( x \)-axis, we can use both the **washer (disc) method** and the **shell method**. Let's proceed step-by-step. ### 1. Identifying the Region \( R \) First, determine the points of intersection between \( y = x \) and \( y = x^{\frac{1}{3}} \): \[ x = x^{\frac{1}{3}} \implies x^3 = x \implies x(x^2 - 1) = 0 \implies x = 0 \text{ or } x = \pm 1 \] Since we're considering the first quadrant (\( x \geq 0 \)), the intersection points are \( x = 0 \) and \( x = 1 \). For \( 0 \leq x \leq 1 \): \[ y = x^{\frac{1}{3}} \geq y = x \] Thus, the region \( R \) is bounded by \( x = 0 \), \( x = 1 \), \( y = x \), and \( y = x^{\frac{1}{3}} \). ### 2. Washer (Disc) Method When revolving around the \( x \)-axis, the volume \( V \) can be computed using: \[ V = \pi \int_{a}^{b} \left[ (y_{\text{upper}})^2 - (y_{\text{lower}})^2 \right] dx \] Here: - Upper curve: \( y = x^{\frac{1}{3}} \) - Lower curve: \( y = x \) - Limits of integration: \( a = 0 \), \( b = 1 \) Thus, \[ V = \pi \int_{0}^{1} \left[ \left( x^{\frac{1}{3}} \right)^2 - x^2 \right] dx = \pi \int_{0}^{1} \left( x^{\frac{2}{3}} - x^2 \right) dx \] Calculating the integral: \[ \int x^{\frac{2}{3}} dx = \frac{3}{5} x^{\frac{5}{3}}, \quad \int x^2 dx = \frac{1}{3} x^3 \] Plugging in the limits: \[ V = \pi \left[ \frac{3}{5}x^{\frac{5}{3}} - \frac{1}{3}x^3 \right]_0^1 = \pi \left( \frac{3}{5} - \frac{1}{3} \right) = \pi \left( \frac{9}{15} - \frac{5}{15} \right) = \pi \left( \frac{4}{15} \right) = \frac{4\pi}{15} \] ### 3. Shell Method For the shell method, it's convenient to integrate with respect to \( y \). Express \( x \) in terms of \( y \): \[ y = x \implies x = y, \quad y = x^{\frac{1}{3}} \implies x = y^3 \] The height of each cylindrical shell is \( x_{\text{right}} - x_{\text{left}} = y - y^3 \), and the radius is \( y \). The volume \( V \) is given by: \[ V = 2\pi \int_{c}^{d} (\text{radius})(\text{height}) dy = 2\pi \int_{0}^{1} y (y - y^3) dy \] Simplifying the integral: \[ V = 2\pi \int_{0}^{1} (y^2 - y^4) dy = 2\pi \left[ \frac{1}{3}y^3 - \frac{1}{5}y^5 \right]_0^1 = 2\pi \left( \frac{1}{3} - \frac{1}{5} \right) = 2\pi \left( \frac{2}{15} \right) = \frac{4\pi}{15} \] ### 4. Conclusion Both methods yield the same result: \[ \boxed{\dfrac{4\,\pi}{15}} \]