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

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Volumes and areas: harder questions Problem set A

Sketch or read the diagram first. Use the calculator for definite integrals and to solve for an unknown limit; give exact answers when the question asks, and 3 significant figures otherwise.

Difficulty: first skills applying them stretch

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 \(a\).
    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 \(b\).
  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.
    2. (b) The shaded region is rotated 360° about the x-axis. Find the volume of the solid.
  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.
    2. (b) Find the volume of the solid formed when the region is rotated 360° about the y-axis.
  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.
    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.
  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.
    2. (b) Find the volume when the region is rotated about the x-axis.
    3. (c) Find the volume when the same region is rotated about the y-axis.
  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 \(x=\frac{\pi}{3}\).
    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).

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.
    2. (b) The region between the curves is rotated 360° about the x-axis. Find the volume of the solid, in 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.
    2. (b) The region is rotated 360° about the y-axis. Find the volume of the solid.
  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.
    2. (b) Use technology to find the area of R.
    3. (c) Use technology to find the volume of the solid formed when R is rotated 360° about the x-axis.
  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.
    2. (b) The cup is rotated 360° about the y-axis. Find the volume of liquid it can hold.
    3. (c) Find the volume of material used to make the cup.

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 \(x^2=\sqrt[3]{(y^2-27)^2}\).
    2. (b) Find the area of R.
    3. (c) Find the volume of the solid formed when R is rotated 360° about the x-axis.
    4. (d) Find the volume of the solid formed when R is rotated 360° about the y-axis.
  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.
    2. (b) Hence show that \(\int_1^a 3\ln x\,dx=3a\ln a-3a+3\).
  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 \((h,r)\).
    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 \(\frac13\pi r^2h\).
    3. (c) Use the formula for a cone with radius 7 cm and height 15 cm.
  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.
    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 \(\frac{\pi}{3}\left(2r^3-3r^2a+a^3\right)\).
    3. (c) Find the volume of the cap when \(r=10\) cm and \(a=6\) cm.
  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.
  6. The graph of \(y=\ln x\) is translated 1 unit to the left.

    1. (a) Write down the equation of the translated curve.
    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.
  7. Consider the curve \(y=2+\sin x\) for \(0\le x\le2\pi\).

    1. (a) Sketch the curve.
    2. (b) The curve is rotated 360° about the x-axis. Show that the volume of the solid is \(\frac{\pi}{2}\int_0^{2\pi}\left(9+8\sin x-\cos 2x\right)dx\).
    3. (c) Find the exact value of the volume.

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