Volumes of Solids of Revolution Lecture #6 a Sphereoid Parabaloid Hyperboloid Whateveroid
Volumes Calculating 3-D Space an Object Occupies Take a cross-sectional slice. Compute the area of the slice. Multiply the area by a tiny thickness. Total all those thin slices with an integral.
Volumes Example # / Part Find the volume of a "cap" of a sphere with radius, "r", and height, "h". This is a view from the top. Here is a sphere. The (red) or darker area is the maximum, projected area of the "cap". The minimum projected area is zero. So, a typical projected area lies between those two extremes.
Volumes Example # / Part If the radius of the sphere is "r", and the height of the cap is "h", then the volume of a thin slice along the vertical axis is "dv". ( ) dv π r y dy Add all of those little volumes,"dv", to obtain the volume that the cap occupies. Since the cap height starts at y = r - h and ends at y = r so too do our limits of integration. V r rh ( ) π r y dy π h r h 3
Volumes Calculation Example # / Part Find the volume of the solid obtained by rotating the region bounded by the given curves about the specified line. y x y x about the x-axis Sketch the 3-D Object. Set the two equations equal to each other to find the point(s) of intersection with each other. x x ( ) x 4 x x x 3 0 x 0 x Points of intersection: ( 0, 0) and (, ). (, ) ( x x 0.5 0 0.5 x
Volumes Calculation Example # / Part The "washers" are the annular rings of thickness, "dx" and width, " x x the axis of rotation. ( ) " along Calculate the area, "A(x)", of an annulus. ( ) ( ) Ax () π x π x The volume of one thin axial slice is "dv". ( ) dv A()dx x π ( x) π x dx Add all of these thin axial slices from the smallest to the largest vale of "x" to obtain the total volume, "V". V 0 Ax () dx π 0 ( x x 4 ) dx 3 π 0
Volumes Calculation Example # 3 / Part Find the volume of the solid obtained by rotating the region under the specified curve from x 0 to x π about the y-axis using the method of "Cylindrical Shells" ( ) y sin x Graph the region to be rotated, showing the little rectangle to be rotated that will form a cylindrical shell. sinx ( ) rx () 0 Sketch a typical cylindrical shell. 0 0.9.8 x The height of the shell is equal to the height, "h(x)" of the little rotated rectangle. ( ) hx () sin x
Volumes Calculation Example # 3 / Part Sketch the annulus of the shell. The area,"da" of the annulus is the product of its circumference, "( π x)", and width, "dx" da π x dx The thickness of the shell is "dx" and its radius, "r" is the distance that the little rectangle is from the axis it is rotated about. In this case, r x. The volume, "dv", of one shell is the product of its height, " sin( x )", and its area, "da". dv sin x ( ) π ( ) x dx
Volumes Example # 3 / Part 3 Add all the volumes of the thin cylindrical shells from x 0 to x π to obtain the total volume, "V", of the solid. V π 0 π x sin( x ) dx π This problem could have been solved by the "disk-washer" approach that we took in the previous example. But, it would be much more labor-intensive.
Parametric Equations & Curves Lecture # 6 b x() t := sin( t + sin() t ) y() t := cos( t + cos() t ) x() t := sin( 3 t) y() t := sin( 4 t) 0.5 yt () 0 yt () 0 0.5 0 xt ( ) 0 xt () This is an example of a "Lissajous" figure. x() t := cos() t y() t := sin( t + sin( 5 t) ) yt () 0 0 xt ( )
PARAMETRIC CURVES: DEFINITION * Generated from separate equations for "x" and "y" coordinates. * They are trajectories in the "x-y" plane that in general fail the "Vertical Line Test". * Equations for "x" and "y" are different but depend on the same independent variable or parameter. * Parametric Equations are more general and therefore more powerful than traditional "y = f(x)" functions.
PARAMETRIC CURVES: EXAMPLE # x() t := t t y() t := t + t ε [ -, 4 ] yt () 5 4 3 0345678 xt () For each value of "t" the separate equations for "x" and "y" are used to calculate their corresponding values so as to produce a point (x,y) on the trajectory. Note that in this case trajectory has starting point at (x(-), y(-)) = (8, -) and ending point at (x(4), y(4)) = (8, 5). So, trajectory is clockwise about this parabola.
PARAMETRIC CURVES: EXAMPLE # * Determine graph of x y 4 3y * Ideal candidate for application of parametric equations. Let "t" be some parameter that takes on all values from - to +. Then if we let "y(t) = t" that forces "y" to take on all values from - to +. That way all possible points on the curve are guaranteed to be generated. y() t := t x() t := t 4 3t yt () 3 3 0 3 3 xt () t ε [ - 3, 3 ]
PARAMETRIC CURVES: EXAMPLE # 3 Describe the motion of a particle with position (x,y) as t varies over the given interval. 0 t π x() t := + cos() t y() t := 3 + sin() t yt () 4 3 This is a circle centered at (,3). 0 0 3 4 xt ( )
PARAMETRIC CURVES: CARTESIAN CONVERSION Given the following parametric equations, find the Cartesian Relationship between "x" & "y". x ( t) + cos ( t ) & y ( t) 3 + sin ( t ) This is possible because of this "Identity": ( sin ( f ( x )) + cos ( f ( x ) ) ) We can use this to our advantage in this case by isolating cos(t) and sin(t) on the left-hand sides of both parametric equations. cos ( t ) ( x ) & sin ( t ) ( y 3) So, x ( ) ( y 3) + which is a circle @ (,3) f ( x) := ( x ) + 3 g ( x) := ( x ) + 3 4 f g ( x) ( x) 3 0 0 3 4 x
PARAMETRIC CURVES: APPLICATION Find the trajectory of the midpoint, "M", of a rod of length, "4". One end of the rod slides along the x-axis and the other end slides along the y-axis. The coordinates of the midpoint are 4 cos( θ) 4 sin( θ) x( θ) y( θ) x θ ( ) cos( θ) := y θ ( ) := sin( θ) ( cos( θ) ) ( sin( θ) ) + x + y x + y ( ) This is a circle of radius "' @ origin. y( θ) 0 x( θ)
INVERSE FUNCTIONS: PARAMETRIC EQUATIONS Find the inverse of " f ", where: f( x) x + sin ( x) Nobody can solve for "x" in terms of "y". But we can cleverly generate "f" and its inverse using parametric equations. Set Set x() t := t and y() t := t + sin () t for " f " v() t := t and u() t t + sin () t := for " f " t ε [ - 0, 0 ] yt () vt () 0 5 0 0 0 5 0 xt (), ut () Here what we obviously did is to successfully graph the original function and its inverse even though we had no clue as to how to solve for "x" in terms of "y". PARAMETRIC EQUATIONS ARE POWERFUL!
TANGENTS TO PARAMETRIC CURVES TANGENTS TO PARAMETRIC CURVES CAN BE FOUND WITHOUT FIRST FINDING THE CARTESIAN EQUATION. xt ():= ft () yt ():= gt () Solve for d dt yt () d dx y = d dx y d dt x d dx y = d dt y d dt x Now let's apply that result to this problem.
TANGENTS TO PARAMETRIC CURVES, Example #, Part # Show that the curve with the following parametric equations has two tangent lines at the origin and find their equations. Illustrate by graphing the curve and its tangents. xt ():= sin() t yt ():= sin( t + sin() t ) d dt y = cos( t + sin() t ) ( + cos() t ) d dt x = cos() t d dx y = cos ( t + sin () t ) ( + cos() t ) cos() t
TANGENTS TO PARAMETRIC CURVES, Example #, Part # But y(x) does a figure-8 every π radians. So at t = 0, we have m = and at t = π, m = 0. Therefore there are two tangent lines at the origin. for the parametric pair: t() x := 0 t() t := sin() t xt ():= sin() t yt ():= sin( t + sin() t ) yt () tt () tt () 0.5 0.5 0 0.5 0.5 xt ()
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