Finding Unknowns from a Volume | CAIE 9709 Pure Mathematics 1 Notes

Pure Mathematics 1

Mathematics9709

Use the volume to find the unknown

When the volume is given, set up the same disc or washer integral, keeping the unknown quantity as a letter. Evaluate the integral before solving the resulting equation. Check the sign and the permitted region at the end.

Worked example

The region under y = k/x from x = 1 to x = 2 is rotated about the x-axis. Its volume is 18π and k > 0. Find k.

The radius is $k/x$ . Squaring introduces $k^2$ , a constant multiplier in an integral with respect to $x$ .

\begin{gathered}18\pi=\pi k^2\int_1^2 x^{-2}\,dx\\=\pi k^2\left[-\frac1x\right]_1^2\\=\frac{\pi k^2}{2}\\k^2=36\end{gathered}

The algebra gives $k=\pm6$ . The condition $k>0$ selects 6. Volume alone could not distinguish a curve above the axis from its reflection below it.

An unknown length belongs in the limit

Your turnPractice

The region under $y=\sqrt{x}$ from 0 to $p$ is rotated about the x-axis. The volume is $8\pi$ , where $p>0$ . Find $p$ .

Show worked answer

\begin{gathered}8\pi=\pi\int_0^p x\,dx=\frac{\pi p^2}{2}\\p^2=16\end{gathered}

Answer

p=4

The negative root fails the domain and the given positive boundary condition.

Your turnSelect the quantity

A region lies between $y=2x$ and $y=x^2$ , for $0\le x\le2$ . Write an integral for its area and another for its volume when rotated about the x-axis. Do not evaluate them.

Show worked answer

The line is above the parabola, and both are non-negative. Area uses the vertical gap; the rotating region has an inner radius $x^2$ .

Answer

\begin{aligned}A&=\int_0^2(2x-x^2)\,dx\\V&=\pi\int_0^2(4x^2-x^4)\,dx\end{aligned}

Use the volume to find the unknown

Worked example

The region under y = k/x from x = 1 to x = 2 is rotated about the x-axis. Its volume is 18π and k > 0. Find k.

The radius is $k/x$ . Squaring introduces $k^2$ , a constant multiplier in an integral with respect to $x$ .

\begin{gathered}18\pi=\pi k^2\int_1^2 x^{-2}\,dx\\=\pi k^2\left[-\frac1x\right]_1^2\\=\frac{\pi k^2}{2}\\k^2=36\end{gathered}

The algebra gives $k=\pm6$ . The condition $k>0$ selects 6. Volume alone could not distinguish a curve above the axis from its reflection below it.

An unknown length belongs in the limit

Your turnPractice

The region under $y=\sqrt{x}$ from 0 to $p$ is rotated about the x-axis. The volume is $8\pi$ , where $p>0$ . Find $p$ .

Show worked answer

\begin{gathered}8\pi=\pi\int_0^p x\,dx=\frac{\pi p^2}{2}\\p^2=16\end{gathered}

Answer

p=4

The negative root fails the domain and the given positive boundary condition.

Your turnSelect the quantity

A region lies between $y=2x$ and $y=x^2$ , for $0\le x\le2$ . Write an integral for its area and another for its volume when rotated about the x-axis. Do not evaluate them.

Show worked answer

The line is above the parabola, and both are non-negative. Area uses the vertical gap; the rotating region has an inner radius $x^2$ .

Answer

\begin{aligned}A&=\int_0^2(2x-x^2)\,dx\\V&=\pi\int_0^2(4x^2-x^4)\,dx\end{aligned}