Area in rectangle-shaped Coordinates

Recall the the area under the graph the a constant function (fleft( x ight)) in between the vertical lines (x = a,) (x = b) deserve to be computed through the identify integral:


where (Fleft( x ight)) is any antiderivative that (fleft( x ight).)

*
Figure 1.

You are watching: Find the area of the region enclosed by the curves y = x and 2x + y2 = 24.

We can prolong the concept of the area under a curve and consider the area that the an ar between 2 curves.

If (fleft( x ight)) and (gleft( x ight)) room two constant functions and also (fleft( x ight) ge gleft( x ight)) top top the closed interval (left< a,b ight>,) then the area between the curve (y = fleft( x ight)) and (y = gleft( x ight)) in this expression is given by


*
Figure 2.

In regards to antiderivatives, the area of region is expressed in the form


dx = Fleft( b ight) - Gleft( b ight) - Fleft( a ight) + Gleft( a ight),>

where (Fleft( x ight)) and (Gleft( x ight)) are antiderivatives the the features (fleft( x ight)) and (gleft( x ight),) respectively.

Note that this area will constantly be non-negative as (fleft( x ight) - gleft( x ight) ge 0) for all (x in left< a,b ight>.)

If there are intersection points, we should break up the interval into several subintervals and determine which curve is better on every subinterval. Then we have the right to determine the area of each an ar by integrating the difference of the larger and the smaller sized function.

Area in Polar Coordinates

Consider the region (OKM) bounded through a polar curve (r = fleft( heta ight)) and two semi-straight lines ( heta =alpha) and also ( heta = eta.)

*
Figure 3.

The area the the polar an ar is provided by



The area that a region between 2 polar curves (r = fleft( heta ight)) and (r = gleft( heta ight)) in the sector (left< alpha ,eta ight>) is express by the integral


d heta .>
*
" width="800" height="779.5">Figure 4.

Area the a region Bounded by a Parametric Curve

Recall that the area under a curve (y = fleft( x ight)) for (fleft( x ight) ge 0) on the interval (left< a,b ight>) deserve to be computed through the integral (intlimits_a^b fleft( x ight)dx.) Suppose now that the curve is characterized in parametric kind by the equations



If the parameter (t) runs between (t_1) and (t_2) where



then the area under the curve is given by the formula



The attributes (xleft( t ight),) (x^primeleft( t ight),) (yleft( t ight)) here are assumed come be continuous on the interval (left< a,b ight>.) as well as that, the function (xleft( t ight),) have to be monotonic on this interval.

*
Figure 5.

If (x = xleft( t ight),) (y = yleft( t ight),) (0 le t le T) room parametric equations that a smooth piecewise closeup of the door curve (C) traversed in the counterclockwise direction and bounding a an ar on the left (Figure (5)), climate the area the the an ar is provided by the following integrals:


dt .>

Solved Problems

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Example 1

Find the area between the curves (y = x^3) and also (y = 3x + 2.)


Example 2

At what value of the parameter (bleft( b gt 1 ight)) the area under the curve (y = x^2) top top the term (left<1,b ight>) is equal to (1?)


Example 3

Find the coordinate of the allude (a) that splits the area under the root role (y = sqrtx) top top the interval (left<0,4 ight>) into equal parts.


Example 4

The region is bounded by the upright lines (x = t), (x = t + fracpi 2), the (x-)axis, and also the curve (y = a + cos x,) whereby (a ge 1.) recognize the value of (t) in ~ which the an ar has the biggest area.


Example 1.

Find the area in between the curve (y = x^3) and also (y = 3x + 2.)


Solution.

First we identify the points of intersection of the curves.

We set (fleft( x ight) = gleft( x ight)) to find the roots:



< Rightarrow x^3 + underbrace x^2 - x^2_0 - 3x - 2 = 0,>
< Rightarrow x^3 + x^2 - x^2 - x - 2x - 2 = 0,>
< Rightarrow x^2left( x + 1 ight) - xleft( x + 1 ight) - 2left( x + 1 ight) = 0,>
< Rightarrow left( x + 1 ight)left( x^2 - x - 2 ight) = 0.>

Solve the quadratic equation:



Thus, the cubic equation has two roots: (x = -1) (of multiplicity (2)) and (x=2.)

The area we wish to calculation is presented in figure (6) below.

*
Figure 6.

We can see from the number that the line (y = 3x + 2) lies over the cubic parabola (y = x^3) on the interval (left< - 1,2 ight>.) Hence, the area that the an ar is offered by


dx = intlimits_ - 1^2 left( 3x + 2 - x^3 ight)dx = left. frac3x^22 + 2x - fracx^44 ight|_ - 1^2 = left( 6 + cancel4 - cancel4 ight) - left( frac32 - 2 - frac14 ight) = frac274.>

Example 2.

At what worth of the parameter (bleft( b gt 1 ight)) the area under the curve (y = x^2) on the term (left<1,b ight>) is equal to (1?)


Solution.

The area under the curve is provided by the integral



*
same to 1." width="800" height="967.1">Figure 7.

Integrating yields:



Hence


4 approx 1.59>

Example 3.

Find the coordinate of the suggest (a) the splits the area under the root role (y = sqrtx) on the interval (left<0,4 ight>) right into equal parts.


Solution.

*
Figure 8.

Equating both areas, we acquire the following:


16 approx 2.52>

Example 4.

The region is bounded by the upright lines (x = t), (x = t + fracpi 2), the (x-)axis, and the curve (y = a + cos x,) whereby (a ge 1.) identify the worth of (t) at which the region has the biggest area.


Solution.

*
through the biggest area." width="800" height="626.6">Figure 9.

See more:
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The area that the region is written in the form



Using the distinction of sines identity



we obtain



The an ar has the biggest area as soon as (cos left( t + fracpi 4 ight) = -1.)

Solving this equation, us find



See much more problems on page 2.


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