Question

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

\(\int_0^4 2 \pi (6 - y)\left( {4y - {y^2}} \right)dy\)

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\((2\pi r \cdot h)\)

\(\int_0^4 2 \pi (6 - y)\left( {4y - {y^2}} \right)dy\)

Solve the expression

Rewrite the expression

\(\begin{aligned}(6 - y)(4y-{y^2}) &\\6 \times4y-6{y^2}-y \times4y-y \times( - {y^2}) &\\24y - 6{y^2} - 4{y^2} + {y^3} &\\24y-10{y^2}+{y^3} &\\{y^3}-10{y^2}+24y\end{aligned}\)

Factor out\(y\)from the expression:

\(y({y^2} - 10y + 24)\)

Evaluate the volume of the solid

\(\int_0^4 {2\pi (6 - y)(4y - {y^2})dy} \)

Substitute\((6 - y)(4y - {y^2}) = y({y^2} - 10y + 24)\)into the expression:

\(\int_0^4 {2\pi y({y^2} - 10y + 24)dy} \)

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:

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

\(V = \int_a^b {2\pi yf(y)dx} \)

Since the rotation is about\(x - \)axis, Substitute\(f(y) = ({y^2} - 10y + 24),a = 0\)and\(b = 4\)into the formula for volume of solid:

\(V = \int_0^4 {2\pi y({y^2} - 10y + 24)dy} \)’

The region under the curve of the function is rotated from\(y = 0\)to\(y = 4\)about the\(x - \)axis.

This can be expressed as:

\(0 \le y \le 4\)

Since the rotation is about the\(x - \)axis, the region under\(x = {y^2} - 10y + 24\)can be expressed as:

\(0 \le x \le {y^2} - 10y + 24\)

Therefore, the solid is formed rotating the regions \(0 \le x \le 4\) and \(0 \le x \le {y^2} - 10y + 24\)about the \(x - \)axis.