Ad tho volume of the solid obtained by rotating the region bounded by \( y=x \) and \( y=\sqrt{x} \) about the line \( x-4 \). aluma \( =\square \) Proview My Answars Bubmil Answes

To find the volume of the solid obtained by rotating the region bounded by \( y=x \) and \( y=\sqrt{x} \) about the line \( x-4 \), we can use the method of cylindrical shells. The volume of the solid can be calculated using the formula: \[ V = 2\pi \int_{a}^{b} (R(x) - r(x))(x-4) dx \] where: - \( R(x) \) is the outer radius of the cylindrical shell, - \( r(x) \) is the inner radius of the cylindrical shell, - \( a \) and \( b \) are the limits of integration. First, let's find the points of intersection between the curves \( y=x \) and \( y=\sqrt{x} \) to determine the limits of integration. We can do this by setting the two equations equal to each other and solving for \( x \): \[ x = \sqrt{x} \] \[ x^2 = x \] \[ x^2 - x = 0 \] \[ x(x-1) = 0 \] So, the points of intersection are \( x=0 \) and \( x=1 \). Now, we can determine the outer radius \( R(x) \) and inner radius \( r(x) \) for the cylindrical shells: - For \( y=x \), the outer radius is \( R(x) = x \) and the inner radius is \( r(x) = \sqrt{x} \). - For \( y=\sqrt{x} \), the outer radius is \( R(x) = \sqrt{x} \) and the inner radius is \( r(x) = x \). Now, we can calculate the volume using the formula: \[ V = 2\pi \int_{0}^{1} (x - \sqrt{x})(x-4) dx \] Let's calculate the integral to find the volume. Evaluate the integral by following steps: - step0: Evaluate using formulas and rules: \(2\pi \times \int_{0}^{1} \left(x-\sqrt{x}\right)\left(x-4\right) dx\) - step1: Evaluate the power: \(2\pi \times \int_{0}^{1} \left(x-x^{\frac{1}{2}}\right)\left(x-4\right) dx\) - step2: Multiply the terms: \(2\pi \times \int_{0}^{1} x^{2}-4x-x^{\frac{3}{2}}+4x^{\frac{1}{2}} dx\) - step3: Evaluate the integral: \(2\pi \times \frac{3}{5}\) - step4: Multiply the terms: \(\frac{6}{5}\pi \) - step5: Multiply: \(\frac{6\pi }{5}\) The volume of the solid obtained by rotating the region bounded by \( y=x \) and \( y=\sqrt{x} \) about the line \( x-4 \) is \( \frac{6\pi}{5} \).