Porisma

Worked solution

Rotate an inverse-tangent region about the y-axis

Find the exact volume of the solid formed when the region bounded by y=tan⁡−1xy=\tan^{-1}x, the y-axis and y=π/4y=\pi/4 is rotated through 360∘360^\circ about the y-axis.

CJC · Lecture · Example 15

Express the radius in terms of y

Rewrite y=tan⁡−1xy=\tan^{-1}x as x=tan⁡yx=\tan y. The selected region lies between the y-axis and this curve for 0≤y≤π/40\le y\le\pi/4. The horizontal asymptotes of the full inverse-tangent graph are y=±π/2y=\pm\pi/2; these are not the region bounds.

How did this part go?

Set up the full-turn volume

V=π∫0π/4x2 dy=π∫0π/4tan⁡2y dyV=\pi\int_0^{\pi/4}x^2\,dy=\pi\int_0^{\pi/4}\tan^2y\,dy. The Teacher highlights the full turn, y-axis, changed radius expression and limits.

How did this part go?

Use the trigonometric identity

From 1+tan⁡2y=sec⁡2y1+\tan^2y=\sec^2y, V=π∫0π/4(sec⁡2y−1) dy=π[tan⁡y−y]0π/4=π(1−π/4)V=\pi\int_0^{\pi/4}(\sec^2y-1)\,dy=\pi[\tan y-y]_0^{\pi/4}=\pi(1-\pi/4) cubic units.

How did this part go?

Compare your sketch

Expand diagram

Use the scrollbars or arrow keys to inspect details.

Inverse-tangent curve with horizontal asymptotes ±π/2. Shade between x=0 and the curve from y=0 toπ/4, then rotate a full turn about the y-axis.
Inverse-tangent curve with horizontal asymptotes ±π/2. Shade between x=0 and the curve from y=0 toπ/4, then rotate a full turn about the y-axis.
Inverse-tangent curve with horizontal asymptotes ±π/2. Shade between x=0 and the curve from y=0 toπ/4, then rotate a full turn about the y-axis.

Common mistakes

The disc radius is x=tan(y), not the inverse tangent of y; the integration variable and bounds must match.

Read the hint