AI HL · AHL 3
0 of 78 skills done
Volumes of revolution about the y-axis
AHL 3.9
Screenshot and video to come
Screenshot and video to come
Screenshot and video to come
Drag the two limits to choose the region. Turn it about the axis and watch each strip sweep out a disc. Then change the number of discs and compare their total with the exact volume.
The solid is drawn as seen from a little to one side, so each circle looks like an ellipse. Each disc takes its radius from the middle of its strip.
Turn the region to see the discs.
The part of the curve \(y=2x^{2}+1\) for \(1\le x\le 3\) is rotated 360° about the y-axis. Find the volume of the solid enclosed by the curve, the y-axis and the horizontal lines through the two ends of the curve. Give the exact value in terms of π, then to 3 significant figures.
Have a go first. Then open the working one line at a time.
\(y=2x^{2}+1\Rightarrow x^{2}=\frac{y-1}{2}\)
You need x², not x, so there is no square root to take.
\(x=1\Rightarrow y=3,\quad x=3\Rightarrow y=19\)
Put each x-limit into the equation of the curve.
\(V=\pi\int_{3}^{19}\frac{y-1}{2}\,dy\)
Integrate with respect to y: the strip is horizontal, with radius x.
From the calculator: \(\pi\int_{3}^{19}\frac{y-1}{2}\,dy = 251\)
Exact value: \(80\pi\).
The part of the curve \(y=\ln(x+1)\) for \(0\le x\le 3\) is rotated 360° about the y-axis. Find the volume of the solid enclosed by the curve, the y-axis and the horizontal lines through the two ends of the curve. Give your answer to 3 significant figures.
Have a go first. Then open the working one line at a time.
\(y=\ln(x+1)\Rightarrow e^{y}=x+1\Rightarrow x=e^{y}-1\)
Undo the ln with e.
\(x^{2}=\left(e^{y}-1\right)^{2}\)
Keep the bracket. Do not square each term separately.
\(x=0\Rightarrow y=\ln 1=0,\quad x=3\Rightarrow y=\ln 4\)
Put each x-limit into the curve.
From the calculator: \(\pi\int_{0}^{\ln 4}\left(e^{y}-1\right)^{2}dy = 9.07\)
3 significant figures. The exact value is \(\pi\left(\frac{3}{2}+2\ln 2\right)\).
The part of the curve \(y=\cos x\) for \(0\le x\le\frac{\pi}{3}\) is rotated 360° about the y-axis. Find the volume of the solid enclosed by the curve, the y-axis and the horizontal lines through the two ends of the curve. Give your answer to 3 significant figures.
Have a go first. Then open the working one line at a time.
\(y=\cos x\Rightarrow x=\cos^{-1}y\)
x is between 0 and π/3, so there is one value of x for each y.
\(x=0\Rightarrow y=1,\quad x=\frac{\pi}{3}\Rightarrow y=0.5\)
Cosine decreases, so the lower y-value comes from the larger x. Put the smaller number at the bottom of the integral.
\(V=\pi\int_{0.5}^{1}\left(\cos^{-1}y\right)^{2}dy\)
Radian mode.
From the calculator: \(\pi\int_{0.5}^{1}\left(\cos^{-1}y\right)^{2}dy = 0.834\)
3 significant figures.
The part of the curve \(y=\frac{1}{x-2}\) for \(3\le x\le 6\) is rotated 360° about the y-axis. Find the volume of the solid enclosed by the curve, the y-axis and the horizontal lines through the two ends of the curve. Give your answer to 3 significant figures.
Have a go first. Then open the working one line at a time.
\(y=\frac{1}{x-2}\Rightarrow x-2=\frac{1}{y}\Rightarrow x=\frac{1}{y}+2\)
Swap, then solve for x.
\(x=3\Rightarrow y=1,\quad x=6\Rightarrow y=\frac{1}{4}\)
The curve is decreasing. The lower limit is 1/4.
\(V=\pi\int_{1/4}^{1}\left(\frac{1}{y}+2\right)^{2}dy\)
Keep the whole bracket for the calculator.
From the calculator: \(\pi\int_{0.25}^{1}\left(\frac{1}{y}+2\right)^{2}dy = 36.3\)
3 significant figures.
The diagram shows the region bounded by the curve \(y=\frac{4}{x}\), the y-axis and the lines \(y=1\) and \(y=8\). (a) Find the area of the region. (b) The region is rotated 360° about the y-axis. Find the exact volume of the solid formed.
Have a go first. Then open the working one line at a time.
\(y=\frac{4}{x}\Rightarrow x=\frac{4}{y}\)
The y-values are already given as the limits.
From the calculator: \(\int_{1}^{8}\frac{4}{y}\,dy = 8.32\)
Area: integrate x. The exact value is 4 ln 8.
\(V=\pi\int_{1}^{8}\frac{16}{y^{2}}\,dy=\pi\left[-\frac{16}{y}\right]_{1}^{8}\)
Volume: integrate πx², with x² = 16/y².
\(V=\pi(-2+16)=14\pi\)
14π is about 44.0.
The region between the curve \(y=\sqrt{x^{2}+1}\), the y-axis and the line \(y=4\) is shown in the diagram. It is rotated 360° about the y-axis. Find the volume of the solid, as a multiple of π.
Have a go first. Then open the working one line at a time.
\(y^{2}=x^{2}+1\Rightarrow x^{2}=y^{2}-1\)
Squaring removes the square root.
\(\text{At }x=0,\ y=\sqrt{1}=1.\ \text{The top is }y=4.\)
The region lies between y = 1 and y = 4.
\(V=\pi\int_{1}^{4}\left(y^{2}-1\right)dy=\pi\left[\frac{y^{3}}{3}-y\right]_{1}^{4}\)
Integrate, then substitute.
\(V=\pi\left(\frac{52}{3}-\left(-\frac{2}{3}\right)\right)=18\pi\)
Check on the calculator: 56.5.
The diagram shows the region bounded by the curve \(x=5-(y-1)^{2}\) and the line \(x=4\). The region is rotated 360° about the y-axis. Find the volume of the solid formed, as a multiple of π.
Have a go first. Then open the working one line at a time.
\(5-(y-1)^{2}=4\Rightarrow (y-1)^{2}=1\Rightarrow y=0\text{ or }y=2\)
These are the y-limits.
\(\text{outer }x=5-(y-1)^{2},\quad\text{inner }x=4\)
The solid has a hole. The line is the inner edge, the curve is the outer edge.
\(V=\pi\int_{0}^{2}\left[\left(5-(y-1)^{2}\right)^{2}-4^{2}\right]dy\)
Subtract the squares of the radii. Do not square the difference.
From the calculator: \(\pi\int_{0}^{2}\left[\left(5-(y-1)^{2}\right)^{2}-16\right]dy = 36.9\)
Exact value: \(\frac{176\pi}{15}\).
A vase is made by rotating the curve \(y=10\ln(x+1)\), for \(0\le y\le 20\), 360° about the y-axis. Lengths are in cm. (a) Find the capacity of the vase in litres. (b) The vase is filled with 250 cm³ of water. Find the depth of the water.
Have a go first. Then open the working one line at a time.
\(y=10\ln(x+1)\Rightarrow x=e^{y/10}-1\)
Rearrange to get x, because the rotation is about the y-axis.
\(V=\pi\int_0^{20}\left(e^{y/10}-1\right)^2dy\)
Limits: the y-values of the base and the top.
From the calculator: \(V=503\text{ cm}^3=0.503\) litres
Divide by 1000 to change cm³ to litres.
Solve \(\pi\int_0^{h}\left(e^{y/10}-1\right)^2dy=250\) for h: \(h=17.3\)
The depth h is the upper limit. The calculator solves for it, and h is less than 20, as it must be.