Question

Each integral represents the volume of a solid. Describe the solid.

\(\int_0^\pi \pi {(2 - sinx)^2}dx\)

Short answer

Therefore, the solid is formed rotating the regions \(0 \le x \le \pi \) and \(0 \le y \le 2 - \sin x\)about the \(y - \)axis.

Step-by-step solution

Given information

The given value is:

When faced with a problem like this, you must choose between the Disk and Washer Methods. inside the integral,\((\pi {r^2})\)or a variant of the Cylindrical Shells Method\((2\pi r \cdot h)\)

\(\int_0^\pi \pi {(2 - sinx)^2}dx\)

Volume of the solid

The volume of a solid obtained by rotating the region under the curve \(y = f(x)\) from \(a\)to \(b\) about the \(y - \)axis is given by:

\(V = \int_a^b {2\pi xf(x)dx} \)

\(\int_a^b {A(x)dx} ,\int_0^\pi {\pi {{(2 - \sin x)}^2}dx} \)

• Comparing the two integrals, the volume of the solid is obtained by the rotating the region under the curve\(A(x) = \pi {(2 - \sin x)^2}\)from\(0\)to\(\pi \)about the\(x - \)axis.
• The region under the curve of the function is rotated from\(x = 0\)to\(x = \pi \)about the\(x - \)axis. This can be expressed as:

\(0 \le x \le \pi \)

• Since the rotation is about the\(x - \)axis, the region under\(y = 2 - \sin x\)can be expressed as:

\(0 \le y \le 2 - \sin x\)

• Therefore, the solid is formed rotating the regions \(0 \le x \le \pi \) and \(0 \le y \le 2 - \sin x\)about the \(x - \)axis.