Hard Further Maths Volumes of revolution Questions

Challenging, exam-style Further Maths Volumes of revolution questions with worked solutions. Stretch yourself on the hardest volumes-of-revolution, x-axis, composite-region, y-axis problems.

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Further Maths34 questionsStep-by-step solutions
Question 1
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}
Question 2
11 markschallenging
A solid is formed by rotating through 2π2\pi radians about the xx-axis the region under the curve y=x2y = x^{2} from x=0x = 0 to x=1x = 1, together with the region under the line y=1y = 1 from x=1x = 1 to x=2x = 2. Which of the following is the exact volume of the solid?
Show worked solution

Worked solution

  1. Split the solid into two parts

    V=V1+V2V=V_{1}+V_{2}

    The two regions do not overlap, so the volumes simply add.

  2. State the volume of revolution formula

    V=π∫y2 dxV=\pi\int y^{2}\,dx

    Rotation about the xx-axis gives discs of radius yy.

  3. Identify the two regions

    0≤y≤x2 (0≤x≤1),0≤y≤1 (1≤x≤2)0\le y\le x^{2}\ (0\le x\le 1),\quad 0\le y\le 1\ (1\le x\le 2)

    One region is bounded by the curve, the other by the straight boundary.

  4. Write the first boundary

    y=x2y=x^{2}

    This generates the first part of the solid.

  5. Square the first radius

    y2=x4y^{2}=x^{4}

    The squared radius is the integrand for the first part.

  6. Set up the first integral

    V1=π∫01(x4) dxV_{1}=\pi\int_{0}^{1}\left(x^{4}\right)\,dx

    Use the limits of the first region.

  7. Evaluate the first integral

    V1=π[x55]01=π5V_{1}=\pi\left[\frac{x^{5}}{5}\right]_{0}^{1}=\frac{\pi}{5}

    This is the volume of the first part of the solid.

  8. Write the second boundary

    y=1y=1

    This generates the second part of the solid.

  9. Square the second radius

    y2=1y^{2}=1

    The squared radius is the integrand for the second part.

  10. Set up the second integral

    V2=π∫12(1) dxV_{2}=\pi\int_{1}^{2}\left(1\right)\,dx

    Use the limits of the second region.

  11. Evaluate the second integral

    V2=π[x]12=πV_{2}=\pi\left[x\right]_{1}^{2}=\pi

    This is the volume of the second part of the solid.

  12. Add the two volumes

    V=π5+πV=\frac{\pi}{5}+\pi

    The total volume is the sum of the two parts.

  13. Simplify the sum

    V=6π5V=\frac{6 \pi}{5}

    Combine the two exact multiples of π\pi over a common denominator.

  14. Check the two parts meet at the join

    x=1x=1

    The two regions share the boundary line at this value, so the solid is continuous.

  15. Confirm both volumes are positive

    V1>0, V2>0V_{1}>0,\ V_{2}>0

    Each rotation produces a genuine solid, so both contributions are positive.

  16. Evaluate numerically as a check

    V≈3.7699V\approx 3.7699

    The decimal value confirms the exact answer is sensible.

  17. Select the option equal to the exact total volume

    6π5\frac{6 \pi}{5}

    Only this option equals V1+V2V_{1}+V_{2}.

Answer
6π5\frac{6 \pi}{5}
Question 3
11 markschallenging
The region RR is bounded by the curve y=9−x2y = \sqrt{9 - x^{2}}, the xx-axis and the lines x=0x = 0 and x=3x = 3. RR is rotated through 2π2\pi radians about the xx-axis. Which of the following integrals gives the exact volume of the solid generated?
Show worked solution

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. Identify the region and its boundaries

    0≤y≤9−x2,0≤x≤30\le y\le \sqrt{9 - x^{2}},\quad 0\le x\le 3

    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=9−x2y=\sqrt{9 - x^{2}}

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

  4. State the radius of a typical disc

    r=y=9−x2r=y=\sqrt{9 - x^{2}}

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

  5. Square the radius

    y2=(9−x2)2=9−x2y^{2}=\left(\sqrt{9 - x^{2}}\right)^{2}=9 - x^{2}

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

  6. Write the definite integral with the given limits

    V=π∫03(9−x2) dxV=\pi\int_{0}^{3} \left(9 - x^{2}\right)\,dx

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

  7. Integrate the squared radius

    ∫(9−x2) dx=−x33+9x+c\int \left(9 - x^{2}\right)\,dx=- \frac{x^{3}}{3} + 9 x+c

    Integrate term by term using the standard results.

  8. Write the result in evaluation-bracket form

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

    The square bracket records the antiderivative ready for substitution.

  9. Substitute the upper limit

    (−x33+9x)x=3=18\left(- \frac{x^{3}}{3} + 9 x\right)_{x=3}=18

    Evaluate the antiderivative at the top of the range.

  10. Substitute the lower limit

    (−x33+9x)x=0=0\left(- \frac{x^{3}}{3} + 9 x\right)_{x=0}=0

    Evaluate the antiderivative at the bottom of the range.

  11. Subtract the lower value from the upper value

    18−0=1818-0=18

    The definite integral is the difference of the two evaluations.

  12. Multiply by π\pi

    V=π×18V=\pi\times 18

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

  13. Check the integration by differentiating

    ddx(−x33+9x)=9−x2\frac{d}{dx}\left(- \frac{x^{3}}{3} + 9 x\right)=9 - x^{2}

    Differentiating the antiderivative must return the integrand.

  14. Write the volume of an elementary disc

    δV≈π(9−x2)2 δx\delta V\approx\pi\left(\sqrt{9 - x^{2}}\right)^{2}\,\delta x

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

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

    V=lim⁡δx→0∑πy2 δxV=\lim_{\delta x\to 0}\sum \pi y^{2}\,\delta x

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

  16. Confirm the volume is positive

    V>0V>0

    A volume must be positive, which is a useful check on the sign.

  17. Select the integral that represents the volume

    π∫03(9−x2)2 dx\pi\int_{0}^{3} \left(\sqrt{9 - x^{2}}\right)^{2}\,dx

    The integrand must be the square of the radius and the factor π\pi must be present.

Answer
π∫03(9−x2)2 dx\pi\int_{0}^{3} \left(\sqrt{9 - x^{2}}\right)^{2}\,dx
Question 4
11 markschallenging
The region RR is bounded by the curve y=sin⁡(x)y = \sin{\left(x \right)}, the xx-axis and the lines x=0x = 0 and x=π2x = \frac{\pi}{2}. RR is rotated through 2π2\pi radians about the xx-axis. Which of the following is the exact volume of the solid generated?
Show worked solution

Worked solution

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

    V=π∫0π2y2 dxV=\pi\int_{0}^{\frac{\pi}{2}} y^{2}\,dx

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

  2. Identify the region and its boundaries

    0≤y≤sin⁡(x),0≤x≤π20\le y\le \sin{\left(x \right)},\quad 0\le x\le \frac{\pi}{2}

    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=sin⁡(x)y=\sin{\left(x \right)}

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

  4. State the radius of a typical disc

    r=y=sin⁡(x)r=y=\sin{\left(x \right)}

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

  5. Square the radius

    y2=(sin⁡(x))2=sin⁡2(x)y^{2}=\left(\sin{\left(x \right)}\right)^{2}=\sin^{2}{\left(x \right)}

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

  6. Write the definite integral with the given limits

    V=π∫0π2(sin⁡2(x)) dxV=\pi\int_{0}^{\frac{\pi}{2}} \left(\sin^{2}{\left(x \right)}\right)\,dx

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

  7. Integrate the squared radius

    ∫(sin⁡2(x)) dx=x2−sin⁡(x)cos⁡(x)2+c\int \left(\sin^{2}{\left(x \right)}\right)\,dx=\frac{x}{2} - \frac{\sin{\left(x \right)} \cos{\left(x \right)}}{2}+c

    Integrate term by term using the standard results.

  8. Write the result in evaluation-bracket form

    V=π[x2−sin⁡(x)cos⁡(x)2]0π2V=\pi\left[\frac{x}{2} - \frac{\sin{\left(x \right)} \cos{\left(x \right)}}{2}\right]_{0}^{\frac{\pi}{2}}

    The square bracket records the antiderivative ready for substitution.

  9. Substitute the upper limit

    (x2−sin⁡(x)cos⁡(x)2)x=π2=π4\left(\frac{x}{2} - \frac{\sin{\left(x \right)} \cos{\left(x \right)}}{2}\right)_{x=\frac{\pi}{2}}=\frac{\pi}{4}

    Evaluate the antiderivative at the top of the range.

  10. Substitute the lower limit

    (x2−sin⁡(x)cos⁡(x)2)x=0=0\left(\frac{x}{2} - \frac{\sin{\left(x \right)} \cos{\left(x \right)}}{2}\right)_{x=0}=0

    Evaluate the antiderivative at the bottom of the range.

  11. Subtract the lower value from the upper value

    π4−0=π4\frac{\pi}{4}-0=\frac{\pi}{4}

    The definite integral is the difference of the two evaluations.

  12. Multiply by π\pi

    V=π×π4V=\pi\times \frac{\pi}{4}

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

  13. Check the integration by differentiating

    ddx(x2−sin⁡(x)cos⁡(x)2)=sin⁡2(x)\frac{d}{dx}\left(\frac{x}{2} - \frac{\sin{\left(x \right)} \cos{\left(x \right)}}{2}\right)=\sin^{2}{\left(x \right)}

    Differentiating the antiderivative must return the integrand.

  14. Write the volume of an elementary disc

    δV≈π(sin⁡(x))2 δx\delta V\approx\pi\left(\sin{\left(x \right)}\right)^{2}\,\delta x

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

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

    V=lim⁡δx→0∑πy2 δxV=\lim_{\delta x\to 0}\sum \pi y^{2}\,\delta x

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

  16. Confirm the volume is positive

    V>0V>0

    A volume must be positive, which is a useful check on the sign.

  17. Select the option equal to the exact volume

    π24\frac{\pi^{2}}{4}

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

Answer
π24\frac{\pi^{2}}{4}
Question 5
11 markschallenging
The region RR is bounded above by the curve y=x+2y = x + 2 and below by the curve y=xy = x, between x=1x = 1 and x=3x = 3. RR is rotated through 2π2\pi radians about the xx-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 xx-axis

    V=π∫13(y12−y22) dxV=\pi\int_{1}^{3}\left(y_{1}^{2}-y_{2}^{2}\right)\,dx

    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

    x≤y≤x+2,1≤x≤3x\le y\le x + 2,\quad 1\le x\le 3

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

  3. Write the outer boundary

    y1=x+2y_{1}=x + 2

    The outer boundary is further from the axis of rotation.

  4. Write the inner boundary

    y2=xy_{2}=x

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

  5. Square the outer radius

    y12=(x+2)2=x2+4x+4y_{1}^{2}=\left(x + 2\right)^{2}=x^{2} + 4 x + 4

    The outer disc has area πy12\pi y_{1}^{2}.

  6. Square the inner radius

    y22=(x)2=x2y_{2}^{2}=\left(x\right)^{2}=x^{2}

    The removed disc has area πy22\pi y_{2}^{2}.

  7. Form the difference of the squares

    y12−y22=4x+4y_{1}^{2}-y_{2}^{2}=4 x + 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∫13(4x+4) dxV=\pi\int_{1}^{3}\left(4 x + 4\right)\,dx

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

  9. Integrate the difference of the squares

    ∫(4x+4) dx=2x2+4x+c\int \left(4 x + 4\right)\,dx=2 x^{2} + 4 x+c

    Integrate term by term.

  10. Write the result in evaluation-bracket form

    V=π[2x2+4x]13V=\pi\left[2 x^{2} + 4 x\right]_{1}^{3}

    The bracket is ready for substitution of the limits.

  11. Substitute the upper limit

    (2x2+4x)x=3=30\left(2 x^{2} + 4 x\right)_{x=3}=30

    Evaluate the antiderivative at the top of the range.

  12. Substitute the lower limit

    (2x2+4x)x=1=6\left(2 x^{2} + 4 x\right)_{x=1}=6

    Evaluate the antiderivative at the bottom of the range.

  13. Subtract the lower value from the upper value

    30−6=2430-6=24

    This is the value of the definite integral.

  14. Multiply by π\pi

    V=π×24V=\pi\times 24

    Restore the factor π\pi from the area of each annulus.

  15. Check the integration by differentiating

    ddx(2x2+4x)=4x+4\frac{d}{dx}\left(2 x^{2} + 4 x\right)=4 x + 4

    Differentiating the antiderivative returns the integrand.

  16. Write the volume of an elementary washer

    δV≈π((x+2)2−(x)2) δx\delta V\approx\pi\left(\left(x + 2\right)^{2}-\left(x\right)^{2}\right)\,\delta x

    A thin slice is an annulus of thickness δx\delta x.

  17. Select the option equal to the exact volume

    24π24 \pi

    Only this option equals π∫(y12−y22) dx\pi\int\left(y_{1}^{2}-y_{2}^{2}\right)\,dx.

Answer
24π24 \pi

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