Let \( R \) be the region bounded by the following curves. Let \( S \) be the solid generated when \( R \) is revolved about the given axis. If possible, find the volume of \( S \) by both the disklwasher and shell methods. Check that your results agree. \( y=(x-5)^{3}-1, x=0, y=7 \); revolved about the \( y \)-axis

To find the volume of the solid generated by revolving the region \( R \) about the \( y \)-axis using the disk method and the shell method, we need to follow these steps: 1. **Disk Method:** - The volume of the solid generated by revolving the region \( R \) about the \( y \)-axis using the disk method is given by the formula: \[ V = \pi \int_{a}^{b} (f(y))^2 \, dy \] - In this case, the function is \( f(x) = (x-5)^3 - 1 \) and the limits of integration are from \( x = 0 \) to \( x = 7 \). 2. **Shell Method:** - The volume of the solid generated by revolving the region \( R \) about the \( y \)-axis using the shell method is given by the formula: \[ V = 2\pi \int_{a}^{b} x \cdot f(x) \, dx \] - In this case, the function is \( f(x) = (x-5)^3 - 1 \) and the limits of integration are from \( x = 0 \) to \( x = 7 \). Let's calculate the volume using both methods and compare the results. Evaluate the integral by following steps: - step0: Evaluate using formulas and rules: \(\int \pi \left(\left(x-5\right)^{3}-1\right)^{2} dx\) - step1: Subtract the terms: \(\int \pi \left(x^{3}-15x^{2}+75x-126\right)^{2} dx\) - step2: Use properties of integrals: \(\pi \times \int \left(x^{3}-15x^{2}+75x-126\right)^{2} dx\) - step3: Calculate: \(\pi \times \int \left(x^{6}-30x^{5}+375x^{4}-2502x^{3}+9405x^{2}-18900x+15876\right) dx\) - step4: Use properties of integrals: \(\pi \left(\int x^{6} dx+\int -30x^{5} dx+\int 375x^{4} dx+\int -2502x^{3} dx+\int 9405x^{2} dx+\int -18900x dx+\int 15876 dx\right)\) - step5: Calculate: \(\pi \times \int x^{6} dx+\pi \times \int -30x^{5} dx+\pi \times \int 375x^{4} dx+\pi \times \int -2502x^{3} dx+\pi \times \int 9405x^{2} dx+\pi \times \int -18900x dx+\pi \times \int 15876 dx\) - step6: Evaluate the integral: \(\frac{\pi x^{7}}{7}+\pi \times \int -30x^{5} dx+\pi \times \int 375x^{4} dx+\pi \times \int -2502x^{3} dx+\pi \times \int 9405x^{2} dx+\pi \times \int -18900x dx+\pi \times \int 15876 dx\) - step7: Evaluate the integral: \(\frac{\pi x^{7}}{7}-5\pi x^{6}+\pi \times \int 375x^{4} dx+\pi \times \int -2502x^{3} dx+\pi \times \int 9405x^{2} dx+\pi \times \int -18900x dx+\pi \times \int 15876 dx\) - step8: Evaluate the integral: \(\frac{\pi x^{7}}{7}-5\pi x^{6}+75\pi x^{5}+\pi \times \int -2502x^{3} dx+\pi \times \int 9405x^{2} dx+\pi \times \int -18900x dx+\pi \times \int 15876 dx\) - step9: Evaluate the integral: \(\frac{\pi x^{7}}{7}-5\pi x^{6}+75\pi x^{5}-\frac{1251\pi x^{4}}{2}+\pi \times \int 9405x^{2} dx+\pi \times \int -18900x dx+\pi \times \int 15876 dx\) - step10: Evaluate the integral: \(\frac{\pi x^{7}}{7}-5\pi x^{6}+75\pi x^{5}-\frac{1251\pi x^{4}}{2}+3135\pi x^{3}+\pi \times \int -18900x dx+\pi \times \int 15876 dx\) - step11: Evaluate the integral: \(\frac{\pi x^{7}}{7}-5\pi x^{6}+75\pi x^{5}-\frac{1251\pi x^{4}}{2}+3135\pi x^{3}-9450\pi x^{2}+\pi \times \int 15876 dx\) - step12: Evaluate the integral: \(\frac{\pi x^{7}}{7}-5\pi x^{6}+75\pi x^{5}-\frac{1251\pi x^{4}}{2}+3135\pi x^{3}-9450\pi x^{2}+15876\pi x\) - step13: Add the constant of integral C: \(\frac{\pi x^{7}}{7}-5\pi x^{6}+75\pi x^{5}-\frac{1251\pi x^{4}}{2}+3135\pi x^{3}-9450\pi x^{2}+15876\pi x + C, C \in \mathbb{R}\) The volume of the solid generated by revolving the region \( R \) about the \( y \)-axis using the disk method is given by the integral: \[ V = \pi \int_{0}^{7} ((x-5)^3 - 1)^2 \, dx \] Now, let's calculate the volume using the shell method. Evaluate the integral by following steps: - step0: Evaluate using formulas and rules: \(\int 2\pi x\left(\left(x-5\right)^{3}-1\right) dx\) - step1: Subtract the terms: \(\int 2\pi x\left(x^{3}-15x^{2}+75x-126\right) dx\) - step2: Multiply the terms: \(\int \left(2\pi x^{4}-30\pi x^{3}+150\pi x^{2}-252\pi x\right) dx\) - step3: Use properties of integrals: \(\int 2\pi x^{4} dx+\int -30\pi x^{3} dx+\int 150\pi x^{2} dx+\int -252\pi x dx\) - step4: Evaluate the integral: \(\frac{2\pi x^{5}}{5}+\int -30\pi x^{3} dx+\int 150\pi x^{2} dx+\int -252\pi x dx\) - step5: Evaluate the integral: \(\frac{2\pi x^{5}}{5}-\frac{15\pi x^{4}}{2}+\int 150\pi x^{2} dx+\int -252\pi x dx\) - step6: Evaluate the integral: \(\frac{2\pi x^{5}}{5}-\frac{15\pi x^{4}}{2}+50\pi x^{3}+\int -252\pi x dx\) - step7: Evaluate the integral: \(\frac{2\pi x^{5}}{5}-\frac{15\pi x^{4}}{2}+50\pi x^{3}-126\pi x^{2}\) - step8: Add the constant of integral C: \(\frac{2\pi x^{5}}{5}-\frac{15\pi x^{4}}{2}+50\pi x^{3}-126\pi x^{2} + C, C \in \mathbb{R}\) The volume of the solid generated by revolving the region \( R \) about the \( y \)-axis using the shell method is given by the integral: \[ V = 2\pi \int_{0}^{7} x \cdot ((x-5)^3 - 1) \, dx \] Now, we can compare the results from both methods to check if they agree.