Because I cannot make sketches occur on these pages, you will need to do this for yourself as part of reading. I’ve inserted diagrams in the sidebar to help a little.
Sketch y= x (5-x).
First we will look at the traditional first step with integration, the area under a curve, in the general form as drawn in the sidebar. You just sketched a negative quadratic, which crosses the x-axis at 0 and 5 and is symmetrical between these points. Draw two vertical lines close together from the x-axis to meet the curve. This is an element of area; I will call this 'a little strip' and I encourage you to draw one whenever you have an integration to do (one of a kind, anyway). Your element has width dx and length (height) of y, so the area of the element is ydx, if we assume that the strip is so narrow we can treat it as a rectangle, not a trapezium. then the integration is ∫ydx = ∫ 5x-x2 dx = [5x2/2 - x3/3] for 0≤x≤5, or 125/2-125/3 = 125/6 square units.
The general formula for area is ∫ydx, but you could draw the strip across the page and do ∫xdy.
Exercise:
1 Find the area under 3x+4y=24, which is a 6-8-10 triangle. Do both ∫ydx, for 0≤x≤8 AND do ∫xdy for 0≤y≤6, just to show that the process works.
2 Find the area under one cycle of the curve y=sin x. i.e. 0≤x≤180º (or π, depending on your choice of units). The integral of sine is negative cosine.
3 Hard: draw x²+y²=r² in the positive quadrant. Draw a little slice of area and try to do ∫ydx.
Do a substitution of x=cosθ, y = sinθ, so dx=-sinθdθ.
Show that this makes ∫ydx become ∫ sin2θdθ and this is true for 0≤θ≤π/2. No, I didn’t lose a minus sign, because I reversed the limits of integration.
Attempt to integrate sin2θ by using cos2θ = 1-2sin2θ. This should show that the area of the quadrant is πr²/4, so the area of the circle is πr². That is a proof of that formula.
Go back to your sketch of y=x(5-x). Imagine we are going to rotate this area around the x-axis, to gain a sort of discus shape. The little strip will become a disk. of radius y and thickness dx. Ther is a poor diagram in teh sidebar. The area of the disk is obviously πy² and the thickness is still dx, so the element of volume is πy²dx and the integration, for 0≤x≤5,
∫ πy²dx = π ∫ x²(5-x)² dx = π ∫ 25x² - 20x³ + x⁴ dx
= π [25x³/3 - 20x⁴/4+ x⁵5] = 5⁵π (1/3 - 1 + 1) = 5⁵ π/3 cubic units.
Exercise:
4 The volume generated by spinning x2+y²=r² around the x-axis for 0≤x≤r is relatively easy. You are looking for 2πr³ / 3 because this is a hemisphere.
5 What about the ellipsoidal equivalent? Try rotating 4x²+9y²=36 around the x-axis.
6 Check that the circle in Q4 could be rotated around the y-axis to produce the same result. Now do the similar rotation for the ellipsoid and the result should be similar (but different, a swap of letters), since the axes of the ellipsoid are different.
Return once more to y=x(5-x). Draw a little section of the arc of the curve. Again, small diagram in teh sidebar. Oh dear, this looks difficult: how long is it? Might it be ds or dl? If it is ds, would ds² = dx² + dy² ? Then the arc length for 0≤x≤5 is ∫ds = ∫ √(dx² + dy²) Which looks formidable. Multiply this by dx/dx and get ∫ √(1+ (dy/dx)²) dx.... really.
Now look at the formula for the curve we’re looking at and see that dy/dx = 5-2x, so ∫ds = ∫(26-20x+4x²)dx which you can integrate and evaluate to 46⅔……. go on, do it.
7 Easy: draw an arc length on the circle and call ds = r dθ. Integrate for θ between 0 and π/2 and multiply the integral by four to have the perimeter of the circle. Proof of formula.
8 Hard: start from ∫ √(1+ (dy/dx)²) dx. Substitute as for Q3, apply some trig rules and gain the same result. This is about the upper limit of C4 on Edexcel, but nowhere near the limit on CIE or MEI, where the necessary integration is a learned standard.
9 Hard: Mechanics and Further Maths: A cricket ball is hit from ground level at 45º across a level pitch and its initial speed is 20m/s. How far does the ball travel horizontally? How far does it travel along the curve? Next time someone asks how far you can throw a ball, or kick one, will you modify your answer?
Sketch y²=x+4, a parabola, for the positive quadrant only, intercepts at 4 and 2. Draw a horizontal strip and treat this as an element with weight; write down the turning moment of the weight measured form the y-axis; the weight is (µg x dy) and the moment is that (µgxdy),(x/2) because the centre of mass is in the middle of the rectangular strip. Substitute expression in y for the x; add these up (integrate); I have
µg/2 ∫x²dy = µg/2 ∫(y²-4)² dy = µg/2 [y⁵/5 - 8y³/3 + 16y] and 0≤y≤2, so 8 8/15 unlessiI made errors. This is the total moment, equal to the total weight (times) position of centre of mass. Now the total area is [y³/3-4y] by integration, so if the centre of mass is at X, then the weight is µg 5⅓ and X is at x=8/5.
10 Find the y co-ordinate of the centre of mass of the parabolic lamina.
DJS 20130220
