Where πy² comes from, areas measured against the y-axis, and what to do when the curve dips below.
Take y = √x from 0 to 4 and rotate it a full turn about the x-axis. Press play. Each thin slice is a disc of radius y, so its volume is πy² times its thickness. Add them all up and you have the integral.
The radius of each disc is just the height of the curve.
About the x-axis: V = ∫ab πy² dx. Each disc has radius y and thickness dx.
About the y-axis: V = ∫ab πx² dy. Each disc has radius x and thickness dy, so everything must be written in terms of y before you start.
Find the volume when y = √x, for 0 ≤ x ≤ 4, is rotated about the x-axis.
Square the whole thing, including any constant. If y = 2x + 1, then y² is 4x² + 4x + 1, not 4x² + 1. Expanding carelessly here is the commonest way to lose a volume question that was otherwise set up perfectly.
Same idea, axes swapped. For the region between x = y² and the y-axis from y = 0 to y = 2:
Check the dx or dy before you integrate. If it says dy, every letter in the integrand must be a y and both limits must be y values. Rearranging the equation first is usually the quickest route.
A plain integral counts area below the axis as negative, so the two can be very different things.
So "find the integral" and "find the area" are different instructions. Sketch it, find where the curve crosses, and split there.
1. Find the volume when y = √x, 0 ≤ x ≤ 4, is rotated about the x-axis. Give it to 2 decimal places.
2. Rotating about the y-axis, the volume is:
3. ∫−11 x³ dx is zero. What is the area enclosed between the curve and the x-axis over that interval?
Writing V = π∫y²dx with the correct limits, before substituting anything, is the setup mark and it is often half the question.
Squaring correctly. An expansion slip here is not recoverable, and it is invisible if you never write y² out as its own line.
If the question says area and the curve crosses the axis, split the integral. If it says integral, do not.
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