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Integration Volumes of revolution
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FM Volumes of revolution II: around y-axis
KUS objectives BAT Find Volumes of revolution using Integration; Rotations around the yAxis Starter: Find these integrals π₯π₯π₯πΆ π₯π₯π₯π₯ π₯π₯π₯ππ₯ π₯π₯π₯
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Notes The volume of revolution formed when x=f(y) is rotated around the y-axis between the y- axis, π=π and π=π is given by ππππ’ππ=π π π π₯ 2 ππ₯ When you use this formula you are integrating with respect to y. So you may need to rearrange functions accordingly
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WB B1 The region R is bounded by the curve π¦ =4 ln π₯ β1, the y-axis, x-axis and the horizontal lines y = 0 and y = 4 Show that the volume of the solid formed when the region is rotated 2Ο radians about the y-axis is 2π π π 2 β1 π₯= π π¦ = π π π¦ 4 π¦ =4 ln π₯ β1, rearranges to π£πππ’ππ=π π π π¦ ππ¦ =π π π π¦ 2 ππ¦ =π π π π¦ =π π π 2 β π 0 =2π π π 2 β QED
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WB B2 The region R is bounded by the curve π¦= ln π₯ the y-axis and the vertical lines π¦=1 and π¦=5 Find the volume of the solid formed when the region is rotated 2Ο radians about the y-axis. Give your answer as a multiple of Ο rearrange π¦= ln π₯ to x = π π¦ π£πππ’ππ=π π π¦ ππ¦ =π 1 5 π 2π¦ ππ¦ = π π 2π¦ 5 1 = π 2 π 10 β π 2
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WB B The area bounded by the curve y= π₯ 2 and the lines π₯=3 and π¦=1 is rotated 2π about the line π¦=1 Find the volume of the solid formed NOW DO Ex 4B transform the graph by the shift f(x)-1 π¦= π₯ 2 becomes π¦= π₯ 2 β1 π£πππ’ππ=π π₯ 2 β ππ₯ =π π₯ 4 β2 π₯ ππ₯ R = π Can you generalise to give a formula for the volume formed when the curve is rotated about line π¦=π transform the graph by the shift f(x)-a so π£πππ’ππ=π π π π π₯ βπ 2 ππ₯
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One thing to improve is β
KUS objectives BAT Find Volumes of revolution using Integration; Rotations around the yAxis self-assess One thing learned is β One thing to improve is β
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