Volumes of revolution Worked Solutions — Further Maths Maths

Fully worked, step-by-step solutions to Further Maths Volumes of revolution questions. See exactly how to solve problems on volumes-of-revolution, x-axis, y-axis, identify.

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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.

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=2xy = 2 x, the xx-axis and the lines x=0x = 0 and x=1x = 1. RR is rotated through 2π2\pi radians about the xx-axis. Find the exact volume of the solid generated.

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=(2x)2=4x2y^{2}=\left(2 x\right)^{2}=4 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[4x33]01=π×43V=\pi\left[\frac{4 x^{3}}{3}\right]_{0}^{1}=\pi\times \frac{4}{3}

    Evaluate the antiderivative at each limit and subtract.

  4. State the exact volume

    V=4π3V=\frac{4 \pi}{3}

    The volume is an exact multiple of π\pi.

Answer
4π3\frac{4 \pi}{3}
Question 3
3 markseasy
The region RR is bounded by the curve y=x2y = x^{2}, the xx-axis and the lines x=0x = 0 and x=2x = 2. RR is rotated through 2π2\pi radians about the xx-axis. Find the exact volume of the solid generated.

Worked solution

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

    V=π∫02y2 dxV=\pi\int_{0}^{2} 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=(x2)2=x4y^{2}=\left(x^{2}\right)^{2}=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[x55]02=π×325V=\pi\left[\frac{x^{5}}{5}\right]_{0}^{2}=\pi\times \frac{32}{5}

    Evaluate the antiderivative at each limit and subtract.

  4. State the exact volume

    V=32π5V=\frac{32 \pi}{5}

    The volume is an exact multiple of π\pi.

Answer
32π5\frac{32 \pi}{5}
Question 4
3 markseasy
The region RR is bounded by the line y=x+1y = x + 1, the xx-axis and the lines x=0x = 0 and x=2x = 2. RR is rotated through 2π2\pi radians about the xx-axis. Find the exact volume of the solid generated.

Worked solution

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

    V=π∫02y2 dxV=\pi\int_{0}^{2} 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+1)2=x2+2x+1y^{2}=\left(x + 1\right)^{2}=x^{2} + 2 x + 1

    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+x2+x]02=π×263V=\pi\left[\frac{x^{3}}{3} + x^{2} + x\right]_{0}^{2}=\pi\times \frac{26}{3}

    Evaluate the antiderivative at each limit and subtract.

  4. State the exact volume

    V=26π3V=\frac{26 \pi}{3}

    The volume is an exact multiple of π\pi.

Answer
26π3\frac{26 \pi}{3}
Question 5
3 markseasy
The region RR is bounded by the curve y=xy = \sqrt{x}, the xx-axis and the lines x=0x = 0 and x=4x = 4. RR is rotated through 2π2\pi radians about the xx-axis. Find the exact volume of the solid generated.

Worked solution

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

    V=π∫04y2 dxV=\pi\int_{0}^{4} 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=xy^{2}=\left(\sqrt{x}\right)^{2}=x

    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[x22]04=π×8V=\pi\left[\frac{x^{2}}{2}\right]_{0}^{4}=\pi\times 8

    Evaluate the antiderivative at each limit and subtract.

  4. State the exact volume

    V=8πV=8 \pi

    The volume is an exact multiple of π\pi.

Answer
8π8 \pi

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