Further Maths Volumes of revolution Practice Questions

Free Further Maths Volumes of revolution practice questions with full step-by-step worked solutions. Covers volumes-of-revolution, x-axis, y-axis, identify. Practise exam-style problems and check your method.

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Further Maths70 questionsStep-by-step solutions
Question 1
3 markseasy
The region RR is bounded by the line y=xy = x, the xx-axis and the lines x=0x = 0 and x=3x = 3. RR is rotated through 2π2\pi radians about the xx-axis. Find the exact volume of the solid generated.
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Worked solution

  1. State the volume of revolution formula for rotation about the xx-axis

    V=π∫03y2 dxV=\pi\int_{0}^{3} y^{2}\,dx

    Rotating the region about the xx-axis sweeps out circular discs of radius yy and thickness dxdx.

  2. Square the radius

    y2=(x)2=x2y^{2}=\left(x\right)^{2}=x^{2}

    The area of the disc is πy2\pi y^{2}, so the squared radius is the integrand.

  3. Integrate and substitute the limits

    V=π[x33]03=π×9V=\pi\left[\frac{x^{3}}{3}\right]_{0}^{3}=\pi\times 9

    Evaluate the antiderivative at each limit and subtract.

  4. State the exact volume

    V=9πV=9 \pi

    The volume is an exact multiple of π\pi.

Answer
9π9 \pi
Question 2
3 markseasy
The region RR is bounded by the line y=x+2y = x + 2, the xx-axis and the lines x=0x = 0 and x=1x = 1. RR is rotated through 2π2\pi radians about the xx-axis. Which of the following is the exact volume of the solid generated?
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Worked solution

  1. State the volume of revolution formula for rotation about the xx-axis

    V=π∫01y2 dxV=\pi\int_{0}^{1} y^{2}\,dx

    Rotating the region about the xx-axis sweeps out circular discs of radius yy and thickness dxdx.

  2. Square the radius

    y2=(x+2)2=x2+4x+4y^{2}=\left(x + 2\right)^{2}=x^{2} + 4 x + 4

    The area of the disc is πy2\pi y^{2}, so the squared radius is the integrand.

  3. Integrate and substitute the limits

    V=π[x33+2x2+4x]01=π×193V=\pi\left[\frac{x^{3}}{3} + 2 x^{2} + 4 x\right]_{0}^{1}=\pi\times \frac{19}{3}

    Evaluate the antiderivative at each limit and subtract.

  4. Select the option equal to the exact volume

    19π3\frac{19 \pi}{3}

    Only this option matches the value of π∫y2 dx\pi\int y^{2}\,dx.

Answer
19π3\frac{19 \pi}{3}
Question 3
5 marksintermediate
The region RR is bounded on the right by the curve x=yx = y and on the left by the curve x=y2x = y^{2}, between y=0y = 0 and y=1y = 1. RR is rotated through 2π2\pi radians about the yy-axis. Which of the following is the exact volume of the solid generated?
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Worked solution

  1. State the washer formula for rotation about the yy-axis

    V=π∫01(x12−x22) dyV=\pi\int_{0}^{1}\left(x_{1}^{2}-x_{2}^{2}\right)\,dy

    Each slice is an annulus: the outer radius sweeps a disc from which the inner disc is removed.

  2. Write the outer boundary

    x1=yx_{1}=y

    The outer boundary is further from the axis of rotation.

  3. Write the inner boundary

    x2=y2x_{2}=y^{2}

    The inner boundary is nearer the axis and creates the hole.

  4. Form the difference of the squares

    x12−x22=−y4+y2x_{1}^{2}-x_{2}^{2}=- y^{4} + y^{2}

    Subtracting gives the area of the annular cross-section divided by π\pi.

  5. Write the definite integral with the given limits

    V=π∫01(−y4+y2) dyV=\pi\int_{0}^{1}\left(- y^{4} + y^{2}\right)\,dy

    The limits are the ends of the region along the axis of rotation.

  6. Integrate and substitute the limits

    V=π[−y55+y33]01=π×215V=\pi\left[- \frac{y^{5}}{5} + \frac{y^{3}}{3}\right]_{0}^{1}=\pi\times \frac{2}{15}

    Evaluate the antiderivative at each limit and subtract.

  7. Select the option equal to the exact volume

    2π15\frac{2 \pi}{15}

    Only this option equals π∫(x12−x22) dy\pi\int\left(x_{1}^{2}-x_{2}^{2}\right)\,dy.

Answer
2π15\frac{2 \pi}{15}
Question 4
8 markshard
The region RR is bounded on the right by the curve x=2x = 2 and on the left by the curve x=y2x = y^{2}, between y=0y = 0 and y=1y = 1. RR is rotated through 2π2\pi radians about the yy-axis. Which of the following is the exact volume of the solid generated?
Show worked solution

Worked solution

  1. State the washer formula for rotation about the yy-axis

    V=π∫01(x12−x22) dyV=\pi\int_{0}^{1}\left(x_{1}^{2}-x_{2}^{2}\right)\,dy

    Each slice is an annulus: the outer radius sweeps a disc from which the inner disc is removed.

  2. Identify the region and its boundaries

    y2≤x≤2,0≤y≤1y^{2}\le x\le 2,\quad 0\le y\le 1

    The region lies between the two curves over the given interval.

  3. Write the outer boundary

    x1=2x_{1}=2

    The outer boundary is further from the axis of rotation.

  4. Write the inner boundary

    x2=y2x_{2}=y^{2}

    The inner boundary is nearer the axis and creates the hole.

  5. Square the outer radius

    x12=(2)2=4x_{1}^{2}=\left(2\right)^{2}=4

    The outer disc has area πx12\pi x_{1}^{2}.

  6. Square the inner radius

    x22=(y2)2=y4x_{2}^{2}=\left(y^{2}\right)^{2}=y^{4}

    The removed disc has area πx22\pi x_{2}^{2}.

  7. Form the difference of the squares

    x12−x22=4−y4x_{1}^{2}-x_{2}^{2}=4 - y^{4}

    Subtracting gives the area of the annular cross-section divided by π\pi.

  8. Write the definite integral with the given limits

    V=π∫01(4−y4) dyV=\pi\int_{0}^{1}\left(4 - y^{4}\right)\,dy

    The limits are the ends of the region along the axis of rotation.

  9. Integrate the difference of the squares

    ∫(4−y4) dy=−y55+4y+c\int \left(4 - y^{4}\right)\,dy=- \frac{y^{5}}{5} + 4 y+c

    Integrate term by term.

  10. Write the result in evaluation-bracket form

    V=π[−y55+4y]01V=\pi\left[- \frac{y^{5}}{5} + 4 y\right]_{0}^{1}

    The bracket is ready for substitution of the limits.

  11. Substitute the upper limit

    (−y55+4y)y=1=195\left(- \frac{y^{5}}{5} + 4 y\right)_{y=1}=\frac{19}{5}

    Evaluate the antiderivative at the top of the range.

  12. Substitute the lower limit

    (−y55+4y)y=0=0\left(- \frac{y^{5}}{5} + 4 y\right)_{y=0}=0

    Evaluate the antiderivative at the bottom of the range.

  13. Select the option equal to the exact volume

    19π5\frac{19 \pi}{5}

    Only this option equals π∫(x12−x22) dy\pi\int\left(x_{1}^{2}-x_{2}^{2}\right)\,dy.

Answer
19π5\frac{19 \pi}{5}
Question 5
11 markschallenging
The region RR is bounded by the curve y=x3y = x^{3}, the yy-axis and the lines y=0y = 0 and y=8y = 8. RR is rotated through 2π2\pi radians about the yy-axis. Which of the following is the exact volume of the solid generated?
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Worked solution

  1. State the volume of revolution formula for rotation about the yy-axis

    V=π∫08x2 dyV=\pi\int_{0}^{8} x^{2}\,dy

    Rotating the region about the yy-axis sweeps out circular discs of radius xx and thickness dydy.

  2. Identify the region and its boundaries

    0≤x≤y3,0≤y≤80\le x\le \sqrt[3]{y},\quad 0\le y\le 8

    The region is bounded by the curve, the axis of rotation and the two given lines.

  3. Write down the equation of the curve

    y=x3y=x^{3}

    This is the boundary that generates the curved surface of the solid.

  4. Rearrange the equation to give xx in terms of yy

    x=y3x=\sqrt[3]{y}

    Rotation about the yy-axis needs the radius expressed as a function of yy.

  5. State the radius of a typical disc

    r=x=y3r=x=\sqrt[3]{y}

    The radius of each disc is the distance from the axis of rotation to the curve.

  6. Square the radius

    x2=(y3)2=y23x^{2}=\left(\sqrt[3]{y}\right)^{2}=y^{\frac{2}{3}}

    The area of the disc is πx2\pi x^{2}, so the squared radius is the integrand.

  7. Write the definite integral with the given limits

    V=π∫08(y23) dyV=\pi\int_{0}^{8} \left(y^{\frac{2}{3}}\right)\,dy

    The limits come from the ends of the region measured along the axis of rotation.

  8. Integrate the squared radius

    ∫(y23) dy=3y535+c\int \left(y^{\frac{2}{3}}\right)\,dy=\frac{3 y^{\frac{5}{3}}}{5}+c

    Integrate term by term using the standard results.

  9. Write the result in evaluation-bracket form

    V=π[3y535]08V=\pi\left[\frac{3 y^{\frac{5}{3}}}{5}\right]_{0}^{8}

    The square bracket records the antiderivative ready for substitution.

  10. Substitute the upper limit

    (3y535)y=8=965\left(\frac{3 y^{\frac{5}{3}}}{5}\right)_{y=8}=\frac{96}{5}

    Evaluate the antiderivative at the top of the range.

  11. Substitute the lower limit

    (3y535)y=0=0\left(\frac{3 y^{\frac{5}{3}}}{5}\right)_{y=0}=0

    Evaluate the antiderivative at the bottom of the range.

  12. Subtract the lower value from the upper value

    965−0=965\frac{96}{5}-0=\frac{96}{5}

    The definite integral is the difference of the two evaluations.

  13. Multiply by π\pi

    V=π×965V=\pi\times \frac{96}{5}

    The factor π\pi comes from the area πr2\pi r^{2} of each disc.

  14. Check the integration by differentiating

    ddy(3y535)=y23\frac{d}{dy}\left(\frac{3 y^{\frac{5}{3}}}{5}\right)=y^{\frac{2}{3}}

    Differentiating the antiderivative must return the integrand.

  15. Write the volume of an elementary disc

    δV≈π(y3)2 δy\delta V\approx\pi\left(\sqrt[3]{y}\right)^{2}\,\delta y

    A thin slice perpendicular to the axis is approximately a cylinder.

  16. Recognise the integral as the limit of a sum of discs

    V=lim⁡δy→0∑πx2 δyV=\lim_{\delta y\to 0}\sum \pi x^{2}\,\delta y

    Adding the discs and letting the thickness tend to zero gives the integral.

  17. Select the option equal to the exact volume

    96π5\frac{96 \pi}{5}

    Only this option matches the value of π∫x2 dy\pi\int x^{2}\,dy.

Answer
96π5\frac{96 \pi}{5}

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