Question

Find the volume of the solid generated by revolving the region bounded by the graphs of the equations about the given lines. \(y=\sqrt{x}, \quad y=0, \quad x=4\) (a) the \(x\) -axis (b) the \(y\) -axis (c) the line \(x=4\) (d) the line \(x=6\)

Short answer

The volumes of the solids obtained by revolving the region bounded by the graphs about (a) the x-axis, (b) the y-axis, (c) line x=4 and (d) line x=6 are (a) \(8\pi\) cubic units, (b) \(\frac{16\pi}{3}\) cubic units, (c) \(16\pi\) cubic units and (d) \(24\pi\) cubic units, respectively.

Step-by-step solution

Apply Disc Method for Rotation about x-axis:

The disc method involves finding the volume of each little disc, then integrating these over the interval of interest. The equation of the region is \(y=\sqrt{x}\), and the bounds are from \(y=0\) to \(y=2\). Therefore, the volume \(V_a\) for (a) when the area is revolved around the x axis is calculated by the formula: \[V_a = \pi \int_0^2 y^2 dx = \pi \int_0^2 xdx\] That simplifies to \(V_a = \pi \left[ \frac{x^2}{2}\right]^2_0 = \frac{16\pi}{2} = 8\pi \, cubic \, units . \]

Apply Disc Method for Rotation about y-axis:

The disc method is also used here, but now with respect to \(y\). Therefore, the volume \(V_b\) for (b) when revolved around the y axis is calculated by the formula: \[V_b = \pi \int_0^2 (4-y^2)dy =\pi \int_0^2 (4-y^2)dy\], which simplifies to \(V_b = \pi[4y-\frac{1}{3}y^3]^2_0 = \frac{16\pi}{3} \, cubic \, units.\]

Use Cylindrical Shells Method for rotation about line \(x=4\):

The cylindrical shell method involves finding the volume of each small shell and then integrating these over the interval. Therefore, the volume \(V_c\) for (c) when revolved around the line \(x=4\) is calculated by the formula: \[V_c = 2\pi \int_0^2 (4-y)(4-\sqrt{y}) dy = 2\pi \int_0^2 (4\sqrt{y}-y) dy\], which simplifies to \(V_c = 2\pi \left[2y^{3/2}-\frac{1}{2}y^2\right]^2_0 = 16\pi\, cubic\, units.\]

Use Cylindrical Shells Method for rotation about line \(x=6\):

Using the cylindrical shell method again, the volume \(V_d\) for (d) when the area is revolved around the line \(x=6\) is calculated by the formula: \[V_d = 2\pi \int_0^2(6-y)(6-\sqrt{y}) dy = 2\pi \int_0^2 (6\sqrt{y}-y) dy\] which simplifies to \(V_d = 2\pi \left[3y^{3/2}-\frac{1}{2}y^2\right]^2_0 = 24\pi\, cubic \, units.\]