Answer:
Part A)
2048π/3 cubic units.
Part B)
8192π/15 units.
Step-by-step explanation:
We are given that R is the finite region bounded by the graphs of functions:
[tex]f(x)=4\sqrt{x}\text{ and } g(x)=x[/tex]
Part A)
We want to find the volume of the solid of revolution obtained when rotating R about the x-axis.
We can use the washer method, given by:
[tex]\displaystyle \pi\int_a^b[R(x)]^2-[r(x)]^2\, dx[/tex]
Where R is the outer radius and r is the inner radius.
Find the points of intersection of the two graphs:
[tex]\displaystyle \begin{aligned} 4\sqrt{x} & = x \\ 16x&= x^2 \\ x^2-16x&= 0 \\ x(x-16) & = 0 \\ x&=0 \text{ and } x=16\end{aligned}[/tex]
Hence, our limits of integration is from x = 0 to x = 16.
Since 4√x ≥ x for all values of x between [0, 16], the outer radius R is f(x) and the inner radius r is g(x). Substitute:
[tex]\displaystyle V=\pi\int_0^{16}(4\sqrt{x})^2-(x)^2\, dx[/tex]
Evaluate:
[tex]\displaystyle \begin{aligned} \displaystyle V&=\pi\int_0^{16}(4\sqrt{x})^2-(x)^2\, dx \\\\ &=\pi\int_0^{16} 16x-x^2\, dx \\\\ &=\pi\left(8x^2-\frac{1}{3}x^3\Big|_{0}^{16}\right)\\\\ &=\frac{2048\pi}{3}\text{ units}^3 \end{aligned}[/tex]
The volume is 2048π/3 cubic units.
Part B)
We want to find the volume of the solid of revolution obtained when rotating R about the y-axis.
First, rewrite each function in terms of y:
[tex]\displaystyle f(y) = \frac{y^2}{16}\text{ and } g(y) = y[/tex]
Solving for the intersection yields y = 0 and y = 16. So, our limits of integration are from y = 0 to y = 16.
The washer method for revolving about the y-axis is given by:
[tex]\displaystyle V=\pi\int_{a}^{b}[R(y)]^2-[r(y)]^2\, dy[/tex]
Since g(y) ≥ f(y) for all y in the interval [0, 16], our outer radius R is g(y) and our inner radius r is f(y). Substitute and evaluate:
[tex]\displaystyle \begin{aligned} \displaystyle V&=\pi\int_{a}^{b}[R(y)]^2-[r(y)]^2\, dy \\\\ &=\pi\int_{0}^{16} (y)^2- \left(\frac{y^2}{16}\right)^2\, dy\\\\ &=\pi\int_0^{16} y^2 - \frac{y^4}{256} \, dy \\\\ &=\pi\left(\frac{1}{3}y^3-\frac{1}{1280}y^5\Bigg|_{0}^{16}\right)\\\\ &=\frac{8192\pi}{15}\text{ units}^3\end{aligned}[/tex]
The volume is 8192π/15 cubic units.
