1 Area calculations. 1.1 Area of an ellipse or a part of it Without using parametric equations

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1 Area calculations. Area of an ellipse or a part of it.. Without using parametric equations We calculate the area in the first quadrant. We start from the standard equation of the ellipse and we put that equation in the form y = f (x). In the first quadrant, x and y are positive. x a + y b = y = b a x a The area is: x a + y b = y = b a x a We first calculate a primitive function a I = x dx We use integration by parts u = a x dv = dx I = a x x x x a x dx = a a x x a x dx a x = a a x x x dx+a dx a x I = a x x+a arcsin x a The area is: A = b [ a x a x+ a arcsin x ] a a = abπ/4 The area of the ellipse is πab... Using parametric equations The parametric equations of an ellipse are x = acost and x = bsint. We calculate the area in the first quadrant. t ranges from π/ to. A = a ydx = bsint.asintdt = π/ ab ( cost)dt = abπ/4 π/ The area of the ellipse is πab.

2 y y=x B..3 Area of a part of an ellipse Wecalculatetheareaofthepartoftheellipsex +y /3 = inthefirstquadrant and enclosed by the x-axis and the line y = x. We work with the parametric equations x = cost y = 3sint We first calculate the t-value corresponding to point B. This is a point on the ellipse where the abscis equals the ordinate. cost = 3sint tant = 3 t = π/6 The area of the part of the ellipse on the right of the dotted line is A = = 3 3 π/6 ydx = π/6 3sint.sintdt ( cost)dt = π The area of the triangle is = 3 8. The requested area is π 3. O A x. Area under a cycloid arc The parametric equations of a cycloid are x = r(t sint) y = r( cos(t)). t is in [,π]. π A = r ( cost)( cost)dt π = r ( cost+cos t)dt = r π = 3πr ( cost+ + cost )dt.3 Area astroid The parametric equations are x = rcos 3 t y = rsin 3 t. We calculate first a quarter of the surface. A = r sin 3 t 3cos t ( sint)dt π/ = 3r cos tsin 4 tdt π/ Now is sin 4 tcos t = (/4)(sintcost).(/)sin t = (/4)sin t.(/)( cost) = (/8)sin t (/8)sin tcost

3 Hence cos tsin 4 tdt = (/8) The area of the astroid is (3/8)πr. sin tdt (/8) sin tcostdt = (/6) ( cos4t)dt (/6) sin tdsint = t 6 64 sin(4t) 48 sin3 t+c.4 Difference of two areas The graph of y = e x/ sinx is a damped oscillation. Plot the graph. We calculate the difference in area between the parts above the x-axis and the parts below the x-axis in [, ]. First, we calculate I = e x sinxdx Using integration by parts. I = e x/ = e x/ cosx e x/ cosxdx [ e x/ sinx+ e x/ sinxdx = e x/ cosx e x sinx I = e x/ (cosx+ sinx) = e x/ (cosx+sinx) I = e x/ (cosx+sinx) ] e x/ sinxdx Ifwenowtakethisbetweentheboundariesand thenwefindapproximately. So, the difference in area between the parts above the x-axis and the parts below the x-axis, in [, ], is approximately..5 Translation of a circle Take only the part of the circle x + y = r above the x-axis. We shift this arc down over a distance a until the area above the x-axis is halved. We want to find this special value of a. The solution is simplified by noting that, in the shifted position, the area above the x-axis is the same as the area between the arc under the x-axis and the x-axis. The equation of the shifted arc is y = r x a We calculate a such that r r ( r x a)dx =

4 r r r x = [ax] r r But the left hand side is the expression for the area of a semicircle and this is equal to πr /. The right hand side is ar. So a = π.r/4..6 An appropriate upper limit Determine t so that the area under the curve y = f(x), in the interval [ln (),t], is equal to π/6. With f(x) = ex We will require that t ln() dx = π/6 ex To calculate the indefinite integral we use the substitution e x = u with u >, then e x = u + and e x dx = udu. So ex dx = udu u(u +) du = u + = atan(u) The equation becomes t ln() = atan e x +C dx = π/6 ex [ atan ex ] t ln() = π/6 atan e x atan() = π/6 atan e x = π/6+π/ ex = tanπ/3 t = ln(4) Volume of a body of revolution y Q. Volume of a truncated cone P b g The radius of the base and upper surface are called, respectively, g and b. The height is h. The truncated cone is created by turning of a rectangular trapezium. The slope of PQ is (g-b)/h and the equation of PQ is y = x(g b)/h+b. We call G and B respectively, the area of base and upper surface. The volume of a truncated cone is h x

5 I = π h ( g b h x+b) dx h = π h ( g b g b h x+b) d( g b h x+b) [ h = π ( g b ] h 3(g b) h x+b)3 h [ = π (g b+b) 3 b 3] 3(g b) = πh g 3 b 3 3 g b = πh 3 (g +gb+b ) = h 3 (πg +πgb+πb ) = h 3 (πg +πb + πg πb ) = h 3 (G+B + GB). Volume of parts of a sphere The sphere has radius r. The volume of the spherical cap with height h is I = π = π r r h (r x )dx [r x x3 3 ] r r h y r-h x = π 3 h (3r h) Now, we make h equal to r, we have the volume 4 3 πr3 of the sphere. h The volume of a segment can easily be calculated as the difference of two spherical caps..3 Volume of a rotating arc of a cycloid The parametric equations of a cycloid are x = r(t sint) y = r( cos(t)). t is in [,π]. The arc rotates around the x-axis in the interval[, πr]. With these boundaries correspond to the t-values en π. I = π = π πr π y dx r ( cost) r( cost)dt π = πr 3 ( cost) 3 dt

6 We first calculate the indefinite integral. I = ( cost) 3 dt = ( 3cost+3cos t cos 3 t)dt = t 3sint+(3/) (+cost)dt cos 3 tdt Say sin t = u = t 3sint+(3/)t+(3/) costdt u+u 3 /3 = t 3sint+(3/)t+(3/4)sint sint+(/3)sin 3 t+c If we then calculate I we find, after simplification, 5π r 3..4 Volume of a rotating astroid The parametric equations are x = rcos 3 t y = rsin 3 t. t changes from π/ to. The volume is I = π = π r π/ y dx r sin 6 t.3r( sint)cos tdt π/ = 6πr 3 cos tsin 7 tdt To calculate the indefinite integral we set cos(t) = u. Then, the volume is easy to find. 3 Length of a curve 3. Example Consider in [-,] the graph of the curve with equation y = 3 ( x ) 3 +3 We calculate the length of the curve. y = x +x +y = +4x +4x 4 +y = +x L = = /3 (+x )dx

7 3. Example Consider in [-,] the graph of the curve with equation (ex +e x ) We calculate the length of the curve. y = (ex e x ) y = 4 (ex +e x ) +y = 4 (ex ++e x ) = 4 (ex +e x ) +y = (ex +e x ) L = = e e (ex +e x )dx 3.3 Length of an arc of a cycloid The parametric equations of a cycloid are x = r(t sint) y = r( cos(t)). t is in [,π]. dy dx = sint cost ( dy dx ) = (sint) ( cost) +( dy dx ) = cost ( cost) = ( cost) +( dy dx ) = cost L = = = = = 8 π π π π cost ( cost)dt costdt sin t dt sin t dt

8 3.4 Length of an arc of an astroid The parametric equations are x = rcos 3 t y = rsin 3 t. t changes from π/ to. We look first at one quarter of the required length. y = dy dx = r.3sin tcostdt r.3cos tsintdt = sint cost +y = cos t +y = cost dx = 3rcos tsintdt If dx is positive then dt is negative. L = = 3r ( 3rcostsint)dt π/ π/ = 3r/ The full length is 6r. costsintdt