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

Volumes and areas: harder questions Problem set A Worked solutions

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 \(x=2\) and \(x=a\) (where \(a>2\)) is rotated about the x-axis. The volume of the solid is \(\frac{\pi}{3}\). Find the value of \(a\).

      • \(V=\pi\int_{2}^{a}\left(\frac1x\right)^2dx=\pi\left[-\frac1x\right]_{2}^{a}=\pi\left(\frac1{2}-\frac1a\right)\)
      • \(\pi\left(\frac1{2}-\frac1a\right)=\frac{\pi}{3}\), so \(\frac1{2}-\frac1a=\frac{1}{3}\)
      • Solve for a: a = 6

    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 \(18\pi\). Find the value of \(b\).

      • \(y=x^2\) gives \(x^2=y\). The limits are y-values: \(y=0\) to \(y=b^2\)
      • \(V=\pi\int_0^{b^2}y\,dy=\frac{\pi b^4}{2}\)
      • \(\frac{\pi b^4}{2}=18\pi\), so \(b^4=36\)
      • Solve for b: b = 2.45 (exact: \(b=\sqrt{6}\))

  2. The diagram shows the curve \(y=\sqrt{x+4}\), which meets the x-axis at the point A. The shaded region is bounded by the curve, the x-axis and the line \(x=6\).

    −4−2246−11234xyy = √(x + 4)A

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

      • At A, \(y=0\): \(x+4=0\)
      • x = −4

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

      • \(V=\pi\int_{-4}^{6}\left(\sqrt{x+4}\right)^2dx=\pi\int_{-4}^{6}(x+4)\,dx\)
      • \(=\pi\left[\frac{x^2}{2}+4x\right]_{-4}^{6}=50 \pi\)
      • V = 157

  3. The diagram shows the region bounded by the y-axis and the curve with equation \(x=9-y^2\).

    246810−4−3−2−11234xyx = 9 − y²

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

      • The curve meets the y-axis where \(9-y^2=0\): \(y=\pm3\)
      • Integrate across, with respect to y: \(A=\int_{-3}^{3}(9-y^2)\,dy\)
      • From the calculator: A = 36

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

      • \(V=\pi\int_{-3}^{3}x^2\,dy=\pi\int_{-3}^{3}(9-y^2)^2\,dy\)
      • From the calculator: V = \(\frac{1296 \pi}{5}\)
      • V = 814

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

    1. (a) The region under the curve \(y=\frac1x\) between \(x=2\) and \(x=5\) (above the x-axis). Find the exact volume.

      • For \(0\le y\le\frac1{5}\), the region reaches from \(x=2\) to \(x=5\): ring volume \(\pi(5^2-2^2)\times\frac1{5}=\frac{21 \pi}{5}\)
      • For \(\frac1{5}\le y\le\frac1{2}\), it reaches from \(x=2\) to \(x=\frac1y\): \(\pi\int\left(\frac1{y^2}-2^2\right)dy=\frac{9 \pi}{5}\)
      • \(V=6 \pi\) = 18.8

    2. (b) The region bounded by the curve \(y=\sqrt{x-3}\), the line \(x=12\) and the x-axis. Find the volume to the nearest integer.

      • \(x=y^2+3\); at \(x=12\), \(y=3\)
      • Cylinder of radius 12 and height 3: \(\pi\times12^2\times3=432 \pi\)
      • Solid between the curve and the y-axis: \(\pi\int_0^{3}(y^2+3)^2dy=\frac{648 \pi}{5}\)
      • \(V=432 \pi-\frac{648 \pi}{5}\) = 950.018, so V = 950

  5. The region between the curve \(y=\sqrt{2x}\), the x-axis and the line \(x=8\) is rotated 360° about an axis.

    246810−112345xyy = √(2x)

    1. (a) Sketch the curve for \(0\le x\le8\). Mark the end point.

      • Passes through the origin and rises steeply, then flattens
      • At \(x=8\): \(y=\sqrt{16}\) = 4.00

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

      • \(V=\pi\int_0^{8}\left(\sqrt{2x}\right)^2dx=\pi\int_0^{8}2x\,dx\)
      • V = \(64 \pi\) = 201

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

      • \(x=\frac{y^2}{2}\), and y runs from 0 to 4.00. The region is under the curve, so subtract from a cylinder of radius 8 and height 4.00
      • \(V=\pi\times8^2\times\sqrt{16}-\pi\int_0^{\sqrt{16}}\left(\frac{y^2}{2}\right)^2dy\)
      • From the calculator: V = \(\frac{1024 \pi}{5}\) = 643

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

    0.3930.7860.51xyy = sin 2xy = cos 2x

    1. (a) The part of the curve \(y=\sin 3x\) between \(x=0\) and \(x=\frac{\pi}{3}\).

      • \(V=\pi\int_0^{\pi/3}\sin^2 3x\,dx\)
      • From the calculator (exact): \(V=\frac{\pi^2}{6}\)

    2. (b) The region bounded by the x-axis and the curves \(y=\sin 2x\) and \(y=\cos 2x\), for \(0\le x\le\frac{\pi}{4}\) (shown).

      • The curves meet where \(\sin 2x=\cos 2x\): \(x=\frac{\pi}{8}\). Below that the lower curve is sine; above it, cosine
      • \(V=\pi\int_0^{\pi/8}\sin^2 2x\,dx+\pi\int_{\pi/8}^{\pi/4}\cos^2 2x\,dx\)
      • From the calculator (exact): \(V=\frac{\pi^2}{8}-\frac{\pi}{4}\) = 0.448

Two curves and technology

  1. A lampshade is moulded in glass. All lengths are in cm. The curves \(y=x^2\) and \(y=\sqrt{8x}\) are shown.

    −0.50.511.522.5312345xyy = √(8x)y = x²O

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

      • \(x^2=\sqrt{8x}\) gives \(x^4=8x\)
      • \(x(x^3-8)=0\): x = 0 or x = 2
      • Points (0, 0) and (2, 4)

    2. (b) The region between the curves is rotated 360° about the x-axis. Find the volume of the solid, in cm³.

      • Between the curves, \(\sqrt{8x}\) is the outer curve
      • \(V=\pi\int_0^{2}\left(8x-x^4\right)dx\)
      • From the calculator: V = \(\frac{48 \pi}{5}\) = 30.2 cm³

  2. The parabola \(y=2x^2\) and the line \(y=6x\) enclose a region, shown shaded.

    12345101520xyy = 2x²y = 6xO

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

      • \(2x^2=6x\) gives \(x(2x-6)=0\)
      • x = 0 or x = 3; y = 18
      • Points (0, 0) and (3, 18)

    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 \(x=\sqrt{\frac{y}{2}}\), line \(x=\frac{y}{6}\)
      • The parabola is the outer curve for \(0\le y\le18\)
      • \(V=\pi\int_0^{18}\left(\frac{y}{2}-\frac{y^2}{36}\right)dy\)
      • From the calculator: V = \(27 \pi\) = 84.8

  3. The diagram shows the curve \(y=(3-x)e^{x/2}\) 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.

    −11234−1123456xyy = (3 − x)e^(x/2)AB

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

      • At A, \(x=0\): \(y=3e^0=3\)
      • At B, \(y=0\): \(3-x=0\) since \(e^{x/2}>0\), so x = 3

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

      • \(\text{Area}=\int_0^{3}(3-x)e^{x/2}\,dx\)
      • From the calculator: area = 7.93

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

      • \(V=\pi\int_0^{3}\left((3-x)e^{x/2}\right)^2dx\)
      • From the calculator: V = 72.8

  4. The cross-section of a ceramic cup is shaded in the diagram. The outer surface is \(y=0.25x^2\) and the inner surface is \(y=2+0.2x^2\), for \(x\ge0\). All lengths are in cm. The two surfaces meet at the rim.

    1234567246810x (cm)y (cm)

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

      • \(0.25x^2=2+0.2x^2\Rightarrow \frac{1}{20}x^2=2\Rightarrow x^2=40\)
      • \(y=0.25\times40=10\), so the height is 10 cm

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

      • Inner surface: \(x^2=\frac{y-2}{0.2}\)
      • \(V=\pi\int_{2}^{10}\frac{y-2}{0.2}\,dy\)
      • From the calculator: V = 503 cm³

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

      • Outer surface: \(x^2=\frac{y}{0.25}\), so the outer solid is \(\pi\int_0^{10}\frac{y}{0.25}\,dy=200\pi=628.3\)
      • Capacity: \(160\pi=502.7\)
      • Material = outer solid − capacity = 628.3 − 502.7 = 125.7 = 126 cm³ (3 s.f.)

Stretch: unknown shapes and proofs

  1. The curve \(y=\sqrt{x^3+27}\) meets the coordinate axes at \(A(-3,0)\) and \(B(0,3 \sqrt{3})\). The region R is bounded by the curve, the x-axis and the y-axis.

    −4−3−2−112−1123456xyy = √(x³ + 27)AB

    1. (a) Show that \(x^2=\sqrt[3]{(y^2-27)^2}\).

      • \(y^2=x^3+27\), so \(x^3=y^2-27\)
      • Raise both sides to the power \(\frac23\): \(x^2=(x^3)^{2/3}=\left(y^2-27\right)^{2/3}\)
      • \(=\sqrt[3]{(y^2-27)^2}\)

    2. (b) Find the area of R.

      • \(\text{Area}=\int_{-3}^0\sqrt{x^3+27}\,dx\)
      • From the calculator: area = 13.1

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

      • \(V=\pi\int_{-3}^0(x^3+27)\,dx\)
      • From the calculator: V = \(\frac{243 \pi}{4}\) = 191

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

      • Use part (a): \(x^2=\sqrt[3]{(y^2-27)^2}\), with y from 0 to \(3 \sqrt{3}\)
      • \(V=\pi\int_0^{3 \sqrt{3}}\sqrt[3]{(y^2-27)^2}\,dy\)
      • From the calculator: V = 109

  2. The diagram shows the curve \(y=3\ln x\). For \(a>1\), the region R is bounded by the curve, the coordinate axes and the line \(y=3\ln a\).

    123456−22468xyy = 3 ln xy = 3 ln a

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

      • Integrate with respect to y. \(y=3\ln x\) gives \(x=e^{y/3}\)
      • \(\text{Area}=\int_0^{3\ln a}e^{y/3}\,dy=\left[3e^{y/3}\right]_0^{3\ln a}=3a-3\)
      • Area = \(3(a-1)\)

    2. (b) Hence show that \(\int_1^a 3\ln x\,dx=3a\ln a-3a+3\).

      • The rectangle with corners at the origin and \((a,3\ln a)\) has area \(3a\ln a\)
      • It is made of R and the area under the curve from 1 to a: \(3a\ln a=\int_1^a 3\ln x\,dx+3(a-1)\)
      • \(\int_1^a 3\ln x\,dx=3a\ln a-3a+3\)

  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 7 cm and height 15 cm.

      • \(V=\frac13\pi\times7^2\times15\)
      • V = \(245 \pi\) = 770 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 \(r=10\) cm and \(a=6\) cm.

      • \(V=\frac{\pi}{3}\left(2(10)^3-3(10)^2(6)+(6)^3\right)\)
      • V = \(\frac{416 \pi}{3}\) = 436 cm³

  5. The diagram shows the graph of \(y=x^4\) 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 \(y=p^4\) and \(y=q^4\).

    1212345678xyy = x⁴

    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.

      • \(\text{Area of A}=\int_p^qx^4\,dx=\frac{q^5-p^5}{5}\)
      • \(x=y^{1/4}\), so \(\text{Area of B}=\int_{p^4}^{q^4}y^{1/4}\,dy=\frac{4}{5}\left(q^5-p^5\right)\)
      • \(\text{A}:\text{B}=\frac{1}{5}:\frac{4}{5}\): the factor \(q^5-p^5\) cancels
      • The ratio is 1 : 4, whatever p and q are

  6. The graph of \(y=\ln x\) is translated 1 unit to the left.

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

      • A translation 1 unit to the left replaces x by \(x+1\)
      • \(y=\ln(x+1)\)

    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.

      • \(y=\ln(x+1)\) gives \(x=e^y-1\). The curve meets the y-axis where \(e^y=1\), at \(y=0\)
      • \(V=\pi\int_{0}^2\left(e^y-1\right)^2dy=\pi\int_{0}^2\left(e^{2y}-2e^y+1\right)dy\)
      • From the calculator (exact): V = \(\frac{\pi}{2}\left(e^4-4e^2+7\right)\) = 50.3

  7. Consider the curve \(y=2+\sin x\) for \(0\le x\le2\pi\).

    1. (a) Sketch the curve.

      • A sin wave with midline \(y=2\), amplitude 1: it runs from \(y=1\) to \(y=3\)
      • Starts at (0, 2) 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 \(\frac{\pi}{2}\int_0^{2\pi}\left(9+8\sin x-\cos 2x\right)dx\).

      • \(V=\pi\int_0^{2\pi}\left(2+\sin x\right)^2dx=\pi\int_0^{2\pi}\left(4+4\sin x+\sin^2x\right)dx\)
      • Use \(\sin^2x=\frac12-\frac12\cos2x\)
      • \(V=\pi\int_0^{2\pi}\left(4+4\sin x+\frac12-\frac12\cos 2x\right)dx=\frac{\pi}{2}\int_0^{2\pi}\left(9+8\sin x-\cos 2x\right)dx\)

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

      • From the calculator (exact): \(V=9 \pi^{2}\)
      • = 88.8

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