Integral calculus
     Applications of the definite integral
      The area of a region in the plane
         The area between the graph of a curve and the coordinate axis
         The area between the graph of a curve and the coordinate axis examples
         The area bounded by a parametric curve
         The area of the sector of a curve in polar coordinates
         The area of the sector of a curve given in Cartesian (or rectangular) coordinates
         The area of the sector of a parametric curve
The area of a region in the plane
The area between the graph of f (x) and the x-axis
If given a continuous nonnegative function  f defined over an interval [a, b] then, the area A enclosed by the curve y = f (x), the vertical lines, x = a and x = b and the x-axis, is defined as
 
The area between the graph of a curve and the coordinate axis examples
Example:   Find the area between the graph of  f (x) = - (1/3)x3 + 3x and the x-axis over the interval defined by two nonnegative successive roots of the given cubic.
Solution:   To find roots we should solve  f (x) = 0,
 
Thus, the area
 
Example:   Find the area between the curve x = - y2 + y + 2 and the y-axis.
Solution:   Since given curve is the parabola whose axis of symmetry is parallel to the x-axis we first calculate its y-intercepts by setting x = 0 to determine the limits of integration,
 
Thus, the area
 
Example:   Find the area of the ellipse  b2x2 + a2y2 = a2b that is symmetrical about the coordinate axes and that intersects the coordinate axes at the points (+ a, 0) and (0, + b).
Solution:   Let write the ellipse in the explicit form y = f (x)
 
Thus, we calculate the area A using the solution of 
the indefinite integral example 48. that is
 
Hence, the area of the ellipse is abp, and by setting b = a we get the area of the circle a2p.
Example:   Find the area A enclosed by the parabola, the line and the x-axis shown in the below figure.
Solution:   First we should find the equation of the parabola and the line using given points.
The parabola is translated in the direction of the x-axis by x0 = 1, so that
and the line
Thus, the area
 
The area bounded by a parametric curve
If a curve is given by the parametric equations x = f (t) and  y = g (t), then the area enclosed by the curve, the vertical lines, x = a and x = b and the x-axis, we obtain from
where,   f (t1)  = a  and  f (t2)  = b,   g (t) > 0  inside  [t1, t2].
Example:   Find the area bounded by the ellipse  x = a cos t,   y = b sin t  ( 0 < t < 2p ).
Solution:    As the ellipse is symmetrical about the coordinate axes we will calculate quarter of the area.
By substituting  x = 0  and  x = into the equation  x = a cos and solving for t, we get the limits of integration  t1p/2  and  t2 = 0 respectively.
Therefore,
so that, the area of the ellipse  A = abp.
The area in polar coordinates
Polar coordinates ( r, q ) locate a point in a plane by means of the length r, of the line joining the point to the origin or pole O, and angle q  swept out by that line from the polar axis.
Recall the area A of a circular sector bounded by two radii and an arc, is defined by the proportion
A : r2 p  = q  : 2p    thus,    A = (1/2) r2q
where r is the length of the radius and q  is the central angle subtended by the arc.
Suppose given a curve in polar coordinates by r = f (q ) or r = r (q ) then, the area of the region bounded by the curve and the radii that correspond to q 1 = a and  q 2 = b, is given by
   
Example:   Find the area of the region enclosed by the lemniscate of Bernoulli whose polar equation is r2 = a2 cos2q, shown in the below figure.
Solution:   As the lemniscate consists of two symmetrical loops meeting at a node, we will calculate quarter of the area lying inside 0 < q  < p/4.
 
Thus, the area of the lemniscate A = a2.
 
The area of the sector of a curve given in Cartesian or rectangular coordinates
We use the formulas for conversion between Cartesian and polar coordinates,
to find the area of the sector bounded by two radii and the arc P0P, of a curve  y = f (x) given in the  
Cartesian coordinates, where x is the value inside the interval [a, b]. Then
therefore,
is the area of the sector bounded by two radii and the arc P0P of a curve  y = f (x).
The area of the sector of a parametric curve
The area of the sector bounded by two radii and the arc P0P of a parametric curve is given by
where the parametric values, t0 and t relate to the endpoints, P0 and P of the arc, respectively.
Example:   Find the area of the sector of the rectangular hyperbola x2 - y2 = 1 enclosed by the x-axis, the arc of the right branch of the hyperbola between the vertex and the point P(cosh t, sinh t), and the line joining this point to the origin, as shown in the below figure.
Solution: Since the hyperbolic functions satisfy the identity  cosh2 t - sinh2 t  = 1  then, the point x = cosh t and  y = sinh t that lies on the hyperbola  x2 - y2 = 1, describes the right branch of the hyperbola as parameter t increases from  - oo  to  + oo.
The point traces the hyperbola passing from negative infinity, through P0 (t = 0) to positive infinity.
 As,       x' (t) = sinh t   and   y' (t) = cosh t,
so that  x(t) y' (t) = cosh2 and   y(t) x' (t) = sinh2 t,
and by using the above formula for the area of the sector
of a parametric curve
 
we get the area of the sector P0OP of the rectangular hyperbola.
 
 Therefore, the parameter (or argument) t, of hyperbolic functions, is twice the area of the sector P0OP,  i.e.,   t = 2A. That is why the inverse hyperbolic functions are also called the area functions.
 
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