Question

question
two rectangles are joined to form the figure below. find the volume of the solid generated
when the figure is rotated about side rw. round your answer to the nearest tenth if
necessary.
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Explanation:

Step1: Calculate the missing side length

The length of \(UV\) is \(9 - 6=3\).

Step2: Divide the figure into two rectangles

Let's consider the figure as two rectangles. One rectangle has dimensions \(10\times6\) and the other has dimensions \(5\times3\).

Step3: Use the formula for the volume of a solid of revolution (using the disk - method, since rotating about an axis, the volume \(V=\pi\int_{a}^{b}[f(x)]^{2}dx\), but for simple shapes, if we rotate a 2 - D shape about an axis, we can use the fact that the volume of a solid of revolution of a shape with area \(A\) rotating about an axis is \(V = A\times(\text{circumference of the path of the centroid})\). Another way is to think of it as the sum of volumes of two cylinders. The formula for the volume of a cylinder is \(V=\pi r^{2}h\).

The area of the first rectangle \(A_1 = 10\times6=60\), and when rotated about \(RW\), the radius of the circular path for all points in this rectangle is the same (since it is adjacent to \(RW\)). The area of the second rectangle \(A_2=5\times3 = 15\).
The volume of the solid of revolution is \(V=\pi\times(A_1 + A_2)\times(\text{distance from the axis of rotation})\). Since we are rotating about \(RW\), we can also use the formula \(V=\pi\sum_{i}r_{i}^{2}h_{i}\).
The first part (rectangle with sides \(10\) and \(6\)): When rotated about \(RW\), its volume \(V_1=\pi\times6^{2}\times10\) (using the formula \(V=\pi r^{2}h\) where \(r = 6\) and \(h = 10\)). The second part (rectangle with sides \(5\) and \(3\)): When rotated about \(RW\), its volume \(V_2=\pi\times3^{2}\times5\) (using the formula \(V=\pi r^{2}h\) where \(r = 3\) and \(h = 5\))

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Answer:

\(1271.7\)