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IB Mathematics: Applications and Interpretation HL · AHL 3: Further Integration · AHL 3.10

Volumes and areas: harder questions Problem set B One question at a time

Try each question on paper first. Open the working a line at a time, then say whether you had it.

Skills

  1. In each part a curve is rotated 360° about an axis and the volume is given. Find the unknown limit.

    1. (a) The part of the curve \(y=\frac{1}{x}\) between \(x=3\) and \(x=a\) (where \(a>3\)) is rotated about the x-axis. The volume of the solid is \(\frac{\pi}{4}\). Find the value of \(a\).

      • \(V=\pi\int_{3}^{a}\left(\frac1x\right)^2dx=\pi\left[-\frac1x\right]_{3}^{a}=\pi\left(\frac1{3}-\frac1a\right)\)
      • \(\pi\left(\frac1{3}-\frac1a\right)=\frac{\pi}{4}\), so \(\frac1{3}-\frac1a=\frac{1}{4}\)
      • Solve for a: a = 12

    2. (b) The part of the parabola \(y=x^2\) between \(x=0\) and \(x=b\) (where \(b>0\)) is rotated about the y-axis. The volume of the solid is \(50\pi\). Find the value of \(b\).

      • \(y=x^2\) gives \(x^2=y\). The limits are y-values: \(y=0\) to \(y=b^2\)
      • \(V=\pi\int_0^{b^2}y\,dy=\frac{\pi b^4}{2}\)
      • \(\frac{\pi b^4}{2}=50\pi\), so \(b^4=100\)
      • Solve for b: b = 3.16 (exact: \(b=\sqrt{10}\))

  2. The diagram shows the curve \(y=\sqrt{x+6}\), which meets the x-axis at the point A. The shaded region is bounded by the curve, the x-axis and the line \(x=5\).

    −6−4−2246−11234xyy = √(x + 6)A

    1. (a) Write down the x-coordinate of A.

      • At A, \(y=0\): \(x+6=0\)
      • x = −6

    2. (b) The shaded region is rotated 360° about the x-axis. Find the volume of the solid.

      • \(V=\pi\int_{-6}^{5}\left(\sqrt{x+6}\right)^2dx=\pi\int_{-6}^{5}(x+6)\,dx\)
      • \(=\pi\left[\frac{x^2}{2}+6x\right]_{-6}^{5}=\frac{121 \pi}{2}\)
      • V = 190

  3. The diagram shows the region bounded by the y-axis and the curve with equation \(x=16-y^2\).

    51015−4−224xyx = 16 − y²

    1. (a) Find the area of the region.

      • The curve meets the y-axis where \(16-y^2=0\): \(y=\pm4\)
      • Integrate across, with respect to y: \(A=\int_{-4}^{4}(16-y^2)\,dy\)
      • From the calculator: \(A=\frac{256}{3}\)

    2. (b) Find the volume of the solid formed when the region is rotated 360° about the y-axis.

      • \(V=\pi\int_{-4}^{4}x^2\,dy=\pi\int_{-4}^{4}(16-y^2)^2\,dy\)
      • From the calculator: V = \(\frac{16384 \pi}{15}\)
      • V = 3430

  4. Each solid is formed by rotating a region 360° about the y-axis.

    1. (a) The region under the curve \(y=\frac1x\) between \(x=3\) and \(x=7\) (above the x-axis). Find the exact volume.

      • For \(0\le y\le\frac1{7}\), the region reaches from \(x=3\) to \(x=7\): ring volume \(\pi(7^2-3^2)\times\frac1{7}=\frac{40 \pi}{7}\)
      • For \(\frac1{7}\le y\le\frac1{3}\), it reaches from \(x=3\) to \(x=\frac1y\): \(\pi\int\left(\frac1{y^2}-3^2\right)dy=\frac{16 \pi}{7}\)
      • \(V=8 \pi\) = 25.1

    2. (b) The region bounded by the curve \(y=\sqrt{x-1}\), the line \(x=10\) and the x-axis. Find the volume to the nearest integer.

      • \(x=y^2+1\); at \(x=10\), \(y=3\)
      • Cylinder of radius 10 and height 3: \(\pi\times10^2\times3=300 \pi\)
      • Solid between the curve and the y-axis: \(\pi\int_0^{3}(y^2+1)^2dy=\frac{348 \pi}{5}\)
      • \(V=300 \pi-\frac{348 \pi}{5}\) = 723.823, so V = 724

  5. The region between the curve \(y=\sqrt{3x}\), the x-axis and the line \(x=12\) is rotated 360° about an axis.

    2468101214246xyy = √(3x)

    1. (a) Sketch the curve for \(0\le x\le12\). Mark the end point.

      • Passes through the origin and rises steeply, then flattens
      • At \(x=12\): \(y=\sqrt{36}\) = 6.00

    2. (b) Find the volume when the region is rotated about the x-axis.

      • \(V=\pi\int_0^{12}\left(\sqrt{3x}\right)^2dx=\pi\int_0^{12}3x\,dx\)
      • V = \(216 \pi\) = 679

    3. (c) Find the volume when the same region is rotated about the y-axis.

      • \(x=\frac{y^2}{3}\), and y runs from 0 to 6.00. The region is under the curve, so subtract from a cylinder of radius 12 and height 6.00
      • \(V=\pi\times12^2\times\sqrt{36}-\pi\int_0^{\sqrt{36}}\left(\frac{y^2}{3}\right)^2dy\)
      • From the calculator: V = \(\frac{3456 \pi}{5}\) = 2170

  6. Find the exact volume of each solid. Each region is rotated 360° about the x-axis.

    0.2620.5240.51xyy = sin 3xy = cos 3x

    1. (a) The part of the curve \(y=\sin 4x\) between \(x=0\) and \(x=\frac{\pi}{4}\).

      • \(V=\pi\int_0^{\pi/4}\sin^2 4x\,dx\)
      • From the calculator (exact): \(V=\frac{\pi^2}{8}\)

    2. (b) The region bounded by the x-axis and the curves \(y=\sin 3x\) and \(y=\cos 3x\), for \(0\le x\le\frac{\pi}{6}\) (shown).

      • The curves meet where \(\sin 3x=\cos 3x\): \(x=\frac{\pi}{12}\). Below that the lower curve is sine; above it, cosine
      • \(V=\pi\int_0^{\pi/12}\sin^2 3x\,dx+\pi\int_{\pi/12}^{\pi/6}\cos^2 3x\,dx\)
      • From the calculator (exact): \(V=\frac{\pi^2}{12}-\frac{\pi}{6}\) = 0.299

Two curves and technology

  1. A paperweight is cut from a block of crystal. All lengths are in mm. The curves \(y=x^2\) and \(y=\sqrt{27x}\) are shown.

    1234246810xyy = √(27x)y = x²O

    1. (a) Find the coordinates of the points where the curves meet.

      • \(x^2=\sqrt{27x}\) gives \(x^4=27x\)
      • \(x(x^3-27)=0\): x = 0 or x = 3
      • Points (0, 0) and (3, 9)

    2. (b) The region between the curves is rotated 360° about the x-axis. Find the volume of the solid, in mm³.

      • Between the curves, \(\sqrt{27x}\) is the outer curve
      • \(V=\pi\int_0^{3}\left(27x-x^4\right)dx\)
      • From the calculator: V = \(\frac{729 \pi}{10}\) = 229 mm³

  2. The parabola \(y=3x^2\) and the line \(y=12x\) enclose a region, shown shaded.

    123451020304050xyy = 3x²y = 12xO

    1. (a) Find the points where the line meets the parabola.

      • \(3x^2=12x\) gives \(x(3x-12)=0\)
      • x = 0 or x = 4; y = 48
      • Points (0, 0) and (4, 48)

    2. (b) The region is rotated 360° about the y-axis. Find the volume of the solid.

      • About the y-axis, write x in terms of y: parabola \(x=\sqrt{\frac{y}{3}}\), line \(x=\frac{y}{12}\)
      • The parabola is the outer curve for \(0\le y\le48\)
      • \(V=\pi\int_0^{48}\left(\frac{y}{3}-\frac{y^2}{144}\right)dy\)
      • From the calculator: V = \(128 \pi\) = 402

  3. The diagram shows the curve \(y=(4-x)e^{x/3}\) for \(x\ge0\). The curve meets the y-axis at A and the x-axis at B. The region R is bounded by the curve and the coordinate axes.

    −112345−11234567xyy = (4 − x)e^(x/3)AB

    1. (a) Find the coordinates of A and B.

      • At A, \(x=0\): \(y=4e^0=4\)
      • At B, \(y=0\): \(4-x=0\) since \(e^{x/3}>0\), so x = 4

    2. (b) Use technology to find the area of R.

      • \(\text{Area}=\int_0^{4}(4-x)e^{x/3}\,dx\)
      • From the calculator: area = 13.1

    3. (c) Use technology to find the volume of the solid formed when R is rotated 360° about the x-axis.

      • \(V=\pi\int_0^{4}\left((4-x)e^{x/3}\right)^2dx\)
      • From the calculator: V = 152

  4. The cross-section of a plastic tumbler is shaded in the diagram. The outer surface is \(y=0.4x^2\) and the inner surface is \(y=1.5+0.3x^2\), for \(x\ge0\). All lengths are in cm. The two surfaces meet at the rim.

    12341234567x (cm)y (cm)

    1. (a) Find the height of the tumbler.

      • \(0.4x^2=1.5+0.3x^2\Rightarrow \frac{1}{10}x^2=1.5\Rightarrow x^2=15\)
      • \(y=0.4\times15=6\), so the height is 6 cm

    2. (b) The tumbler is rotated 360° about the y-axis. Find the volume of liquid it can hold.

      • Inner surface: \(x^2=\frac{y-1.5}{0.3}\)
      • \(V=\pi\int_{1.5}^{6}\frac{y-1.5}{0.3}\,dy\)
      • From the calculator: V = 106 cm³

    3. (c) Find the volume of material used to make the tumbler.

      • Outer surface: \(x^2=\frac{y}{0.4}\), so the outer solid is \(\pi\int_0^{6}\frac{y}{0.4}\,dy=45\pi=141.4\)
      • Capacity: \(33.75\pi=106.0\)
      • Material = outer solid − capacity = 141.4 − 106.0 = 35.3 cm³ (3 s.f.)

Stretch: unknown shapes and proofs

  1. The curve \(y=\sqrt{x^3+64}\) meets the coordinate axes at \(A(-4,0)\) and \(B(0,8)\). The region R is bounded by the curve, the x-axis and the y-axis.

    −5−4−3−2−1122468xyy = √(x³ + 64)AB

    1. (a) Show that \(x^2=\sqrt[3]{(y^2-64)^2}\).

      • \(y^2=x^3+64\), so \(x^3=y^2-64\)
      • Raise both sides to the power \(\frac23\): \(x^2=(x^3)^{2/3}=\left(y^2-64\right)^{2/3}\)
      • \(=\sqrt[3]{(y^2-64)^2}\)

    2. (b) Find the area of R.

      • \(\text{Area}=\int_{-4}^0\sqrt{x^3+64}\,dx\)
      • From the calculator: area = 26.9

    3. (c) Find the volume of the solid formed when R is rotated 360° about the x-axis.

      • \(V=\pi\int_{-4}^0(x^3+64)\,dx\)
      • From the calculator: V = \(192 \pi\) = 603

    4. (d) Find the volume of the solid formed when R is rotated 360° about the y-axis.

      • Use part (a): \(x^2=\sqrt[3]{(y^2-64)^2}\), with y from 0 to \(8\)
      • \(V=\pi\int_0^{8}\sqrt[3]{(y^2-64)^2}\,dy\)
      • From the calculator: V = 297

  2. The diagram shows the curve \(y=2\ln x\). For \(a>1\), the region R is bounded by the curve, the coordinate axes and the line \(y=2\ln a\).

    123456−22468xyy = 2 ln xy = 2 ln a

    1. (a) Find, in terms of \(a\), the area of R.

      • Integrate with respect to y. \(y=2\ln x\) gives \(x=e^{y/2}\)
      • \(\text{Area}=\int_0^{2\ln a}e^{y/2}\,dy=\left[2e^{y/2}\right]_0^{2\ln a}=2a-2\)
      • Area = \(2(a-1)\)

    2. (b) Hence show that \(\int_1^a 2\ln x\,dx=2a\ln a-2a+2\).

      • The rectangle with corners at the origin and \((a,2\ln a)\) has area \(2a\ln a\)
      • It is made of R and the area under the curve from 1 to a: \(2a\ln a=\int_1^a 2\ln x\,dx+2(a-1)\)
      • \(\int_1^a 2\ln x\,dx=2a\ln a-2a+2\)

  3. A cone has base radius \(R\) and height \(H\). Place its vertex at the origin with its axis along the x-axis.

    1. (a) Write down the equation of the line through the origin and the point \((H,R)\).

      • Gradient \(\frac{R-0}{H-0}=\frac{R}{H}\) and it passes through the origin
      • \(y=\frac{R}{H}x\)

    2. (b) The part of this line between \(x=0\) and \(x=H\) is rotated 360° about the x-axis. Show that the volume of the cone formed is \(\frac13\pi R^2H\).

      • \(V=\pi\int_0^{H}y^2\,dx=\pi\int_0^{H}\frac{R^2}{H^2}x^2\,dx\)
      • \(=\pi\frac{R^2}{H^2}\left[\frac{x^3}{3}\right]_0^{H}=\pi\frac{R^2}{H^2}\times\frac{H^3}{3}\)
      • \(=\frac13\pi R^2H\)

    3. (c) Use the formula for a cone with radius 9 cm and height 4 cm.

      • \(V=\frac13\pi\times9^2\times4\)
      • V = \(108 \pi\) = 339 cm³

  4. A spherical cap is the part of a sphere on one side of a plane. Take a circle with centre the origin and radius r.

    1. (a) Write down the equation of the circle.

      • A point (x, y) on the circle is a distance r from the origin: \(x^2+y^2=r^2\)

    2. (b) The region bounded by the upper half of the circle, the x-axis and the line \(x=a\) (where \(0<a<r\)) is rotated 360° about the x-axis. Show that the volume of the cap formed is \(\frac{\pi}{3}\left(2r^3-3r^2a+a^3\right)\).

      • On the upper half, \(y^2=r^2-x^2\). The region lies between \(x=a\) and \(x=r\)
      • \(V=\pi\int_a^r\left(r^2-x^2\right)dx=\pi\left[r^2x-\frac{x^3}{3}\right]_a^r\)
      • \(=\pi\left(r^3-\frac{r^3}{3}-r^2a+\frac{a^3}{3}\right)=\frac{\pi}{3}\left(2r^3-3r^2a+a^3\right)\)

    3. (c) Find the volume of the cap when \(r=8\) cm and \(a=5\) cm.

      • \(V=\frac{\pi}{3}\left(2(8)^3-3(8)^2(5)+(5)^3\right)\)
      • V = \(63 \pi\) = 198 cm³

  5. The diagram shows the graph of \(y=x^5\) for \(x\ge0\), and two regions. Region A is under the curve between \(x=p\) and \(x=q\), where \(0<p<q\). Region B is bounded by the curve, the y-axis and the lines \(y=p^5\) and \(y=q^5\).

    121234567891011xyy = x⁵

    1. (a) Show that the ratio \(\text{area of A}:\text{area of B}\) does not depend on \(p\) or \(q\), and find the ratio.

      • \(\text{Area of A}=\int_p^qx^5\,dx=\frac{q^6-p^6}{6}\)
      • \(x=y^{1/5}\), so \(\text{Area of B}=\int_{p^5}^{q^5}y^{1/5}\,dy=\frac{5}{6}\left(q^6-p^6\right)\)
      • \(\text{A}:\text{B}=\frac{1}{6}:\frac{5}{6}\): the factor \(q^6-p^6\) cancels
      • The ratio is 1 : 5, whatever p and q are

  6. The graph of \(y=\ln x\) is translated 2 units to the left.

    1. (a) Write down the equation of the translated curve.

      • A translation 2 units to the left replaces x by \(x+2\)
      • \(y=\ln(x+2)\)

    2. (b) The region R is bounded by the translated curve, the y-axis and the line \(y=2\). Find the exact volume of the solid formed when R is rotated 360° about the y-axis.

      • \(y=\ln(x+2)\) gives \(x=e^y-2\). The curve meets the y-axis where \(e^y=2\), at \(y=\ln 2\)
      • \(V=\pi\int_{\ln 2}^2\left(e^y-2\right)^2dy=\pi\int_{\ln 2}^2\left(e^{2y}-4e^y+4\right)dy\)
      • From the calculator (exact): V = \(\frac{\pi}{2}\left(e^4-8e^2+28-8\ln 2\right)\) = 28.2

  7. Consider the curve \(y=3+\cos x\) for \(0\le x\le2\pi\).

    1. (a) Sketch the curve.

      • A cos wave with midline \(y=3\), amplitude 1: it runs from \(y=2\) to \(y=4\)
      • Starts at (0, 4) and completes one period at x = 2π

    2. (b) The curve is rotated 360° about the x-axis. Show that the volume of the solid is \(\frac{\pi}{2}\int_0^{2\pi}\left(19+12\cos x+\cos 2x\right)dx\).

      • \(V=\pi\int_0^{2\pi}\left(3+\cos x\right)^2dx=\pi\int_0^{2\pi}\left(9+6\cos x+\cos^2x\right)dx\)
      • Use \(\cos^2x=\frac12+\frac12\cos2x\)
      • \(V=\pi\int_0^{2\pi}\left(9+6\cos x+\frac12+\frac12\cos 2x\right)dx=\frac{\pi}{2}\int_0^{2\pi}\left(19+12\cos x+\cos 2x\right)dx\)

    3. (c) Find the exact value of the volume.

      • From the calculator (exact): \(V=19 \pi^{2}\)
      • = 188

jmaths.xyz · AI HL · AHL 3.10 Harder volumes and areas of revolution