Llse the shall method to write and evaluate the definite integral that reprecer \[ \begin{array}{l}y=x^{3 / 2} \\ y=27 \\ x\end{array}=0 \]

To set up the definite integral using the shell method for the given curves \( y = x^{3/2} \) and \( y = 27 \), we need to follow these steps: 1. **Identify the region of integration**: We are looking for the area between the curve \( y = x^{3/2} \) and the line \( y = 27 \) from \( x = 0 \) to the point where they intersect. 2. **Find the intersection points**: Set \( x^{3/2} = 27 \) to find the value of \( x \) at which the two curves intersect. 3. **Set up the shell method integral**: The shell method involves integrating with respect to \( y \). The formula for the shell method is: \[ V = 2\pi \int_{a}^{b} (radius)(height) \, dy \] where the radius is the distance from the axis of rotation (in this case, the y-axis) and the height is the difference between the outer and inner functions. 4. **Evaluate the integral**. Let's go through these steps in detail. ### Step 1: Find the intersection points Set \( x^{3/2} = 27 \): \[ x^{3/2} = 27 \] To solve for \( x \), raise both sides to the power of \( \frac{2}{3} \): \[ x = 27^{2/3} \] Calculating \( 27^{2/3} \): \[ 27^{2/3} = (3^3)^{2/3} = 3^2 = 9 \] So, the curves intersect at \( x = 9 \). ### Step 2: Set up the shell method integral The radius of the shell is \( y \) (the distance from the y-axis), and the height of the shell is given by the difference between the line and the curve: \[ \text{Height} = 27 - y^{2/3} \] Thus, the integral becomes: \[ V = 2\pi \int_{0}^{27} y \cdot (27 - y^{2/3}) \, dy \] ### Step 3: Evaluate the integral Now we will evaluate the integral: \[ V = 2\pi \int_{0}^{27} (27y - y^{5/3}) \, dy \] Let's calculate this integral. 1. Calculate the integral of \( 27y \): \[ \int 27y \, dy = \frac{27y^2}{2} \] 2. Calculate the integral of \( y^{5/3} \): \[ \int y^{5/3} \, dy = \frac{y^{8/3}}{8/3} = \frac{3}{8}y^{8/3} \] Now, we can combine these results and evaluate from \( 0 \) to \( 27 \): \[ V = 2\pi \left[ \left( \frac{27y^2}{2} - \frac{3}{8}y^{8/3} \right) \bigg|_{0}^{27} \right] \] Calculating at the upper limit \( y = 27 \): \[ \frac{27(27^2)}{2} - \frac{3}{8}(27^{8/3}) \] Calculating \( 27^2 = 729 \) and \( 27^{8/3} = (27^{2/3})^4 = 9^4 = 6561 \): \[ = \frac{27 \cdot 729}{2} - \frac{3}{8} \cdot 6561 \] Calculating \( \frac{27 \cdot 729}{2} = \frac{19683}{2} \) and \( \frac{3 \cdot 6561}{8} = \frac{19683}{8} \): \[ = \frac{19683}{2} - \frac{19683}{8} \] Finding a common denominator (8): \[ = \frac{19683 \cdot 4}{8} - \frac{19683}{8} = \frac{19683 \cdot 3}{8} = \frac{59049}{8} \] Thus, the volume \( V \) is: \[ V = 2\pi \cdot \frac{59049}{8} = \frac{118098\pi}{8} = \frac{59049\pi}{4} \] ### Final Answer The volume of the solid formed by rotating the region around the y-axis is: \[ V = \frac{59049\pi}{4} \]