sketch and describe the solid that is produced when the region enclosed by the given equations is rotated…

Explanation:

Step1: Identify the region

The region is bounded by (x = 0), (y=0) and (y=x + 3). It is a right - triangle in the first - quadrant with vertices ((0,0)), ((0,3)) and ((- 3,0)) (but we are in the first - quadrant, so we consider the part where (x\geq0) and (y\geq0)). When rotated about the (x) - axis, it forms a cone.

Step2: Use the disk method formula

The formula for the volume (V) of a solid of revolution using the disk method when rotating about the (x) - axis is (V=\pi\int_{a}^{b}[f(x)]^{2}dx). Here, (y = f(x)=x + 3), and the limits of integration are found by setting (y = 0) and (x=0). The line (y=x + 3) intersects (x = 0) at (y=3) and (y = 0) at (x=-3). Since we are in the first - quadrant, we integrate from (x = 0) to (x=-3). The formula becomes (V=\pi\int_{0}^{-3}(x + 3)^{2}dx). Expand ((x + 3)^{2}=x^{2}+6x + 9).

Step3: Integrate term - by - term

(\int(x^{2}+6x + 9)dx=\frac{x^{3}}{3}+3x^{2}+9x+C).

Step4: Evaluate the definite integral

(V=\pi\left[\frac{x^{3}}{3}+3x^{2}+9x\right]_{0}^{-3}). [ \begin{align*} V&=\pi\left(\frac{(-3)^{3}}{3}+3(-3)^{2}+9(-3)-0\right)\ &=\pi\left(-9 + 27-27\right)\ &=9\pi \end{align*} ]

Answer:

The solid is a cone. The volume of the solid is (9\pi) cubic units.