Question

In Exercises 61 and \(62,\) sketch the solid that has the given description in cylindrical coordinates. $$ 0 \leq \theta \leq \pi / 2,0 \leq r \leq 2,0 \leq z \leq 4 $$

Short answer

The resulting solid is a quarter cylindrical column with radius \(2\) and height \(4\), located in the first quadrant and extending above the x-y plane.

Step-by-step solution

Understanding the Constraints

Here, the three dimensions to consider are represented by the cylindrical coordinates \(r\), \(\theta\) and \(z\). \nFor \(\theta\), the range is from \(0\) to \(\pi/2\), which means the shape spans from the positive x-axis to the positive y-axis. \nFor \(r\), the range is from \(0\) to \(2\), so the shape extends from the origin to a distance of 2 units. \nLastly, for \(z\), the range is from \(0\) to \(4\), indicating that the shape extends upward from the x-y plane to a height of 4 units.

Sketching the Solid

Begin by sketching the x-y plane in two dimensions. Considering the constraints on \(\theta\), draw lines from the origin to the positive y and x-axes. Next, take into account the constraints on \(r\), and draw a circle with radius \(2\), centered at the origin, and bounded by the lines just created. This forms a sector which is the base of the solid. Finally, considering the \(z\) constraints, the shape extends from this base up to a height of \(4\) units. Thus, the final solid is a quarter cylindrical column with radius \(2\) and height \(4\).