a solid has as its base the region in the xy - plane bounded by the graphs of $x^{2}=16y$ and $y = 2$. every…

Explanation:

Step1: Solve for (x) in terms of (y)

Given (x^{2}=16y), then (x = \pm4\sqrt{y}). The length of the base of the rectangle (in the (xy -)plane) at a given (y) is (l=4\sqrt{y}-(- 4\sqrt{y})=8\sqrt{y}).

Step2: Determine the height of the rectangle

The height (h) of the rectangle is twice the length of the side in the (xy -)plane. So (h = 2\times8\sqrt{y}=16\sqrt{y}).

Step3: Set up the integral for the volume

The cross - sectional area (A(y)) of the rectangle is (A(y)=l\times h). Substituting (l = 8\sqrt{y}) and (h = 16\sqrt{y}), we get (A(y)=(8\sqrt{y})\times(16\sqrt{y})=128y). The solid is bounded by (y = 0) and (y = 2). Using the formula for the volume of a solid with known cross - sectional area (V=\int_{a}^{b}A(y)dy), where (a = 0), (b = 2) and (A(y)=128y). So (V=\int_{0}^{2}128y\ dy).

Step4: Evaluate the integral

Using the power rule for integration (\int y^{n}dy=\frac{y^{n + 1}}{n+1}+C) ((n\neq - 1)), we have (\int_{0}^{2}128y\ dy=128\times\frac{y^{2}}{2}\big|{0}^{2}). First, (128\times\frac{y^{2}}{2}=64y^{2}). Then (64y^{2}\big|{0}^{2}=64\times(2^{2}-0^{2})=64\times4 = 256).

Answer:

256