find the volume of the solid generated by revolving the region bounded by the given equations about the y…

find the volume of the solid generated by revolving the region bounded by the given equations about the y - axis\n$x = y^{3/2}$, $x = 0$, and $y = 7$\nthe volume of the solid is $\\frac{4\\pi}{5}$ cubic units.\n(type an exact answer, using $\\pi$ as needed)

find the volume of the solid generated by revolving the region bounded by the given equations about the y - axis\n$x = y^{3/2}$, $x = 0$, and $y = 7$\nthe volume of the solid is $\\frac{4\\pi}{5}$ cubic units.\n(type an exact answer, using $\\pi$ as needed)

Answer

Explanation:

Step1: Use the disk method formula

The formula for the volume (V) of a solid of revolution about the (y -)axis using the disk method is (V=\pi\int_{a}^{b}[x(y)]^{2}dy). Here, (x = y^{3/2}), (a = 0), and (b = 7). So, (V=\pi\int_{0}^{7}(y^{3/2})^{2}dy).

Step2: Simplify the integrand

((y^{3/2})^{2}=y^{3}). Then the integral becomes (V=\pi\int_{0}^{7}y^{3}dy).

Step3: Integrate

Using the power - rule for integration (\int y^{n}dy=\frac{y^{n + 1}}{n+1}+C) ((n\neq - 1)), we have (\int y^{3}dy=\frac{y^{4}}{4}+C). So, (V=\pi\left[\frac{y^{4}}{4}\right]_{0}^{7}).

Step4: Evaluate the definite integral

(V=\pi\left(\frac{7^{4}}{4}-\frac{0^{4}}{4}\right)). Since (7^{4}=2401), then (V=\frac{2401\pi}{4}) cubic units.

Answer:

(\frac{2401\pi}{4})