use the shell method to find the volume of a solid obtained by rotating the region b about the x…

Explanation:

Step1: Find the radius and height of the shell

When using the Shell Method for rotation about the (x -)axis, for a horizontal shell, the radius (r=y) and the height (h) is found by solving (y = x^{2}+b) for (x). So (x=\sqrt{y - b}) (since (x\geq0) in the region shown). The limits of integration for (y) are from (y = b) to (y=a^{2}+b). Given (a = 3) and (b = 5), the limits are from (y = 5) to (y=3^{2}+5=14).

Step2: Set up the Shell - Method formula

The formula for the Shell Method when rotating about the (x -)axis is (V=2\pi\int_{c}^{d}r\cdot h\ dy). Here, (r = y), (h=\sqrt{y - 5}), (c = 5), and (d = 14). So (V=2\pi\int_{5}^{14}y\sqrt{y - 5}\ dy).

Step3: Use substitution

Let (u=y - 5), then (y=u + 5) and (dy=du). When (y = 5), (u = 0); when (y = 14), (u = 9). The integral becomes (2\pi\int_{0}^{9}(u + 5)\sqrt{u}\ du=2\pi\int_{0}^{9}(u^{\frac{3}{2}}+5u^{\frac{1}{2}})\ du).

Step4: Integrate term - by - term

Using the power rule (\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C(n\neq - 1)), we have: (\int(u^{\frac{3}{2}}+5u^{\frac{1}{2}})\ du=\frac{u^{\frac{3}{2}+1}}{\frac{3}{2}+1}+5\frac{u^{\frac{1}{2}+1}}{\frac{1}{2}+1}+C=\frac{2}{5}u^{\frac{5}{2}}+\frac{10}{3}u^{\frac{3}{2}}+C).

Step5: Evaluate the definite integral

(2\pi\left[\frac{2}{5}u^{\frac{5}{2}}+\frac{10}{3}u^{\frac{3}{2}}\right]_{0}^{9}) (=2\pi\left(\frac{2}{5}(9)^{\frac{5}{2}}+\frac{10}{3}(9)^{\frac{3}{2}}-0\right)) Since (9^{\frac{1}{2}} = 3), (9^{\frac{3}{2}}=27), (9^{\frac{5}{2}}=243) (=2\pi\left(\frac{2\times243}{5}+\frac{10\times27}{3}\right)) (=2\pi\left(\frac{486}{5}+90\right)) (=2\pi\left(\frac{486 + 450}{5}\right)) (=2\pi\times\frac{936}{5}=\frac{1872\pi}{5})

Answer:

(\frac{1872\pi}{5})