Rotation About the y-axis | CAIE 9709 Pure Mathematics 1 Notes

Pure Mathematics 1

Mathematics9709

Measure the radius horizontally from the y-axis

About the y-axis, a horizontal strip becomes a disc. The radius is the horizontal distance $|x|$ , and the thickness is measured along $y$ . Use $dy$ and y-values for the limits.

Use y-limits 2 and 5, with x² = y. The ellipses represent circular cross-sections viewed at an angle.

V=\pi\int_a^b x^2\,dy

Express $x^2$ entirely in $y$ before integrating. As before, the region must reach the rotation axis for the disc formula to apply directly.

You may already know x²

Worked example

Rotate the region between y = x², the y-axis, y = 2 and y = 5 about the y-axis.

Use the branch $x\ge0$ . The equation already gives $x^2=y$ , so no square-root calculation is needed.

\begin{gathered}V=\pi\int_2^5y\,dy\\=\pi\left[\frac{y^2}{2}\right]_2^5\\=\frac{21\pi}{2}\end{gathered}

Its length along the rotation axis is $5-2=3$ . Radii run from $\sqrt2$ to $\sqrt5$ , so the volume lies between $6\pi$ and $15\pi$ , the corresponding cylinder volumes.

Your turnPractice

Rotate the region between $y=\sqrt{2x+1}$ , the y-axis and $y=3$ about the y-axis. Find the exact volume.

Show worked answer

The limits are $y=1,3$ . Rearranging gives $x=(y^2-1)/2$ , which must be squared.

\begin{gathered}\int_1^3(y^4-2y^2+1)\,dy\\=\left[\frac{y^5}{5}-\frac{2y^3}{3}+y\right]_1^3\\=\frac{496}{15}\\V=\frac\pi4\times\frac{496}{15}=\frac{124\pi}{15}\end{gathered}

A circle centred at the origin satisfies $x^2+y^2=R^2$ , where $R$ is its radius. Rotating its right-hand semicircular region about the y-axis forms a sphere. At height $y$ , the disc radius satisfies $x^2=R^2-y^2$ .

Your turnShow that

Use integration, not the supplied sphere formula, to show that this sphere has volume $4\pi R^3/3$ , with $R>0$ .

Show worked answer

The bottom and top are $y=-R$ and $y=R$ . Treat $R$ as a constant.

\begin{gathered}V=\pi\int_{-R}^R(R^2-y^2)\,dy\\=\pi\left[R^2y-\frac{y^3}{3}\right]_{-R}^R\\=\pi\left(\frac{2R^3}{3}+\frac{2R^3}{3}\right)\\=\frac{4\pi R^3}{3}\end{gathered}

This agrees with MF19. Agreement is a check; quoting MF19 alone would not be an integration proof.

Measure the radius horizontally from the y-axis

About the y-axis, a horizontal strip becomes a disc. The radius is the horizontal distance $|x|$ , and the thickness is measured along $y$ . Use $dy$ and y-values for the limits.

Use y-limits 2 and 5, with x² = y. The ellipses represent circular cross-sections viewed at an angle.

V=\pi\int_a^b x^2\,dy

Express $x^2$ entirely in $y$ before integrating. As before, the region must reach the rotation axis for the disc formula to apply directly.

You may already know x²

Worked example

Rotate the region between y = x², the y-axis, y = 2 and y = 5 about the y-axis.

Use the branch $x\ge0$ . The equation already gives $x^2=y$ , so no square-root calculation is needed.

\begin{gathered}V=\pi\int_2^5y\,dy\\=\pi\left[\frac{y^2}{2}\right]_2^5\\=\frac{21\pi}{2}\end{gathered}

Its length along the rotation axis is $5-2=3$ . Radii run from $\sqrt2$ to $\sqrt5$ , so the volume lies between $6\pi$ and $15\pi$ , the corresponding cylinder volumes.

Your turnPractice

Rotate the region between $y=\sqrt{2x+1}$ , the y-axis and $y=3$ about the y-axis. Find the exact volume.

Show worked answer

The limits are $y=1,3$ . Rearranging gives $x=(y^2-1)/2$ , which must be squared.

\begin{gathered}\int_1^3(y^4-2y^2+1)\,dy\\=\left[\frac{y^5}{5}-\frac{2y^3}{3}+y\right]_1^3\\=\frac{496}{15}\\V=\frac\pi4\times\frac{496}{15}=\frac{124\pi}{15}\end{gathered}

Your turnShow that

Use integration, not the supplied sphere formula, to show that this sphere has volume $4\pi R^3/3$ , with $R>0$ .

Show worked answer

The bottom and top are $y=-R$ and $y=R$ . Treat $R$ as a constant.

\begin{gathered}V=\pi\int_{-R}^R(R^2-y^2)\,dy\\=\pi\left[R^2y-\frac{y^3}{3}\right]_{-R}^R\\=\pi\left(\frac{2R^3}{3}+\frac{2R^3}{3}\right)\\=\frac{4\pi R^3}{3}\end{gathered}

This agrees with MF19. Agreement is a check; quoting MF19 alone would not be an integration proof.