Area Bounded by Curves

Introduction
In calculus, one of the fundamental
applications of integration is finding the area
enclosed by curves. This technique, often
referred to as definite integration, allows us to
calculate the exact area under a curve within
specific limits. In this assignment, we will
delve into three specific scenarios:
* Area Bounded by a Parabola and an Oblique
Line:
* Area Bounded by a Circle and a Line:
* Area Bounded by Two Parabolas:
1. Area Bounded by a Parabola and an
Oblique Line
Understanding the Concept:
When a parabola and an oblique line
intersect, they can enclose a region of finite
area. To calculate this area, we need to
determine the points of intersection and then
integrate the difference between the two
functions over the appropriate interval.
Steps Involved:
* Find the Points of Intersection:
* Set the equations of the parabola and the
line equal to each other and solve for x.
* This will give you the x-coordinates of the
points where the curves intersect.
* Determine the Limits of Integration:
* The limits of integration will be the x-
coordinates of the points of intersection found
in step 1.
* Set Up the Integral:
* The area, A, can be calculated using the
definite integral:
A = ∫[f(x) – g(x)] dx

Where:
* f(x) is the equation of the curve that is
above the other curve in the interval.
 * g(x) is the equation of the curve that is
below the other curve in the interval.
* The limits of integration are the x-
coordinates of the intersection points.
* Evaluate the Integral:
* Use integration techniques to evaluate the
definite integral.
Example:
Find the area bounded by the parabola y = x²
and the line y = 2x + 3.
Solution:
* Find the Points of Intersection:
* Set x² = 2x + 3.
* Solve the quadratic equation to get x = -1
and x = 3.
* Determine the Limits of Integration:
* The limits of integration are -1 and 3.
* Set Up the Integral:
* The line y = 2x + 3 is above the parabola
y = x² in the interval [-1, 3].
 * So, the integral becomes:
A = ∫[-1, 3] [(2x + 3) – x²] dx

* Evaluate the Integral:

* Integrate the expression and evaluate it at
the limits of integration.
* The final answer is 32/3 square units.
2. Area Bounded by a Circle and a Line
Understanding the Concept:
When a line intersects a circle, it can divide
the circle into two segments. The area of the
segment can be calculated by finding the area
of the sector and subtracting the area of the
triangle formed by the chord and the radii to
the endpoints of the chord.
Steps Involved:
* Find the Points of Intersection:
* Solve the equations of the circle and the
line simultaneously to find the points of
intersection.
* Determine the Angle of the Sector:
* Use trigonometry to find the angle
between the radii to the points of intersection.
* Calculate the Area of the Sector:
* Use the formula for the area of a sector:
A_sector = (θ/2) * r², where θ is the angle in
radians and r is the radius of the circle.
* Calculate the Area of the Triangle:
* Use the formula for the area of a triangle:
A_triangle = (1/2) * base * height.
* Find the Area of the Segment:
* Subtract the area of the triangle from the
area of the sector.
Example:
Find the area of the segment cut off from the
circle x² + y² = 16 by the line y = x.
Solution:
* Find the Points of Intersection:
 * Solve the equations x² + y² = 16 and y =
x simultaneously to get the points of
intersection: (2√2, 2√2) and (-2√2, -2√2).
* Determine the Angle of the Sector:
* The angle between the radii to the points
of intersection is π/2 radians.
* Calculate the Area of the Sector:
* The radius of the circle is 4.
* The area of the sector is (π/4) * 16 = 4π.
* Calculate the Area of the Triangle:
* The base of the triangle is 4√2 and the
height is 4√2.
* The area of the triangle is (1/2) * 4√2 *
4√2 = 16.
* Find the Area of the Segment:
* The area of the segment is 4π – 16 square
units.
3. Area Bounded by Two Parabolas
Understanding the Concept:
When two parabolas intersect, they can
enclose a region of finite area. To calculate
this area, we need to determine the points of
intersection and then integrate the difference
between the two functions over the
appropriate interval.
Steps Involved:
* Find the Points of Intersection:
* Set the equations of the two parabolas
equal to each other and solve for x.
* Determine the Limits of Integration:
* The limits of integration will be the x-
coordinates of the points of intersection.
* Set Up the Integral:
* The area, A, can be calculated using the
definite integral:
A = ∫[f(x) – g(x)] dx

Where:
* f(x) is the equation of the parabola that is
above the other parabola in the interval.
 * g(x) is the equation of the parabola that
is below the other parabola in the interval.
* The limits of integration are the x-
coordinates of the intersection points.
* Evaluate the Integral:
* Use integration techniques to evaluate the
definite integral.
Example:
Find the area bounded by the parabolas y = x²
and y = 4 – x².
Solution:
* Find the Points of Intersection:
* Set x² = 4 – x².
* Solve the equation to get x = ±√2.
* Determine the Limits of Integration:
* The limits of integration are -√2 and √2.
* Set Up the Integral:
* The parabola y = 4 – x² is above the
parabola y = x² in the interval [-√2, √2].
* So, the integral becomes:
 A = ∫[-√2, √2] [(4 – x²) – x²] dx

* Evaluate the Integral:

* Integrate the expression and evaluate it at
the limits of integration.
* The final answer is 16√2/3 square units.
Q1. Sketch a graph of the two functions
between 0 and π.
Note that cos x is above sin x only between x
= 0 and x = π/4
so
we have 5 times integral from 0 to π/4 of cos
x dx - sin x dx
cos x dx --> sin x 0 to π/4 = .5√2
sin x dx = -cos x 0 to π/4= -.5√2 +1
cos x dx - sin x dx --> √2 - 1
times 5 = 5√2 - 5
or about 7.07 - 5 = 2.07
Answer:
Well, since you like to play with curves, let's
find some area under them!

To find the area between the curves y =

5(cosx) and y = 5(sinx) for 0 < x < √, we'll
need to integrate. However, the graphs of
these functions keep playing their little dance
and intersecting each other.

So, let's figure out the points where they

intersect. We know that y = 5(cosx) and y =
5(sinx), so when they are equal to each other,
we get:

5(cosx) = 5(sinx).

Dividing both sides by 5, we have:

cosx = sinx.
Now, we can solve this equation. But instead
of getting too serious with math, let's take a
clownish shortcut. We can use our knowledge
that sin(π/4) = cos(π/4) = √(2)/2.

So, we can say that x = π/4 is one of the

points where the graphs intersect.

Now, we'll need to find the other point of

intersection. We know that the graphs repeat
every 2*π, so we can add 2*π to our previous
solution:

x = π/4 + 2*π.

Now, let's set up the integral to find the area

between the curves:

∫[0, π/4 + 2*π] (5(cosx) - 5(sinx)) dx.

But hey, we're still not done clowning around!
Let's simplify this a bit:

∫[0, π/4 + 2*π] (5cosx - 5sinx) dx.

Now, we can integrate term by term:

∫[0, π/4 + 2*π] 5cosx dx - ∫[0, π/4 + 2*π]

5sinx dx.

Integrating cosx gives us sinx, and integrating

sinx gives us -cosx:

[5sinx] [0, π/4 + 2*π] - [-5cosx] [0, π/4 + 2*π].

Plugging in the limits of integration:

[5sin(π/4 + 2*π) - 5sin(0)] - [-5cos(π/4 + 2*π) -

(-5cos(0))].
Simplifying a bit:

[5sin(π/4 + 2*π) - 5sin(0)] - [-5cos(π/4 + 2*π)

+ 5].

But wait, we're not done yet! Let's appreciate

the symmetry of sine and cosine:

[5(sin(π/4)cos(2*π) + cos(π/4)sin(2*π)) -
5sin(0)] - [-5(cos(π/4)cos(2*π) -
sin(π/4)sin(2*π)) + 5].

Now, let's simplify further:

[5(sin(π/4)(1) + cos(π/4)(0)) - 5sin(0)] - [-

5(cos(π/4)(1) - sin(π/4)(0)) + 5].

Fortunately, sin(0) = 0, so we can simplify

even more:
[5(sin(π/4) - 0)] - [-5(cos(π/4) - 0) + 5].

Now, let's use our clownish knowledge again:

sin(π/4) = cos(π/4) = √(2)/2:

[5(√(2)/2) - 0] - [-5(√(2)/2) - 0 + 5].

Simplifying further:

[5√(2)/2] - [-5√(2)/2 + 5].

Finally, let's do the math:

[5√(2)/2] - [-5√(2)/2 + 5] = 5√(2)/2 + 5√(2)/2 -

5 = 10√(2)/2 - 5 = 5√(2) - 5.
And voila! The area under the curve y =
5(cosx) and above y = 5(sinx) for 0 < x < π is
5√(2) - 5

Conclusion
In this exploration of finding areas bounded
by curves, we’ve delved into three
fundamental scenarios: areas bounded by a
parabola and an oblique line, a circle and a
line, and two parabolas. Through the
application of integral calculus, we’ve
developed a systematic approach to
determine these areas, involving the following
key steps:
* Identifying the Region:
* Visualize the region enclosed by the given
curves, paying attention to their intersections
and relative positions.
* Sketch a graph to aid in understanding the
region’s boundaries.
* Determining the Limits of Integration:
 * Find the x-coordinates of the points of
intersection between the curves.
* These intersection points will serve as the
lower and upper limits of the definite integral.
* Setting Up the Integral:
* Identify the curve that forms the upper
boundary of the region and the curve that
forms the lower boundary.
* Subtract the equation of the lower curve
from the equation of the upper curve to
obtain the integrand.
* Set up the definite integral with the
appropriate limits of integration.
* Evaluating the Integral:
* Apply the fundamental theorem of
calculus to evaluate the definite integral.
* This involves finding the antiderivative of
the integrand and evaluating it at the upper
and lower limits of integration.
* The difference between these evaluations
yields the area of the bounded region.
Key Considerations and Applications
* Symmetry: If the region exhibits symmetry,
it can significantly simplify the calculations.
By exploiting symmetry, we can often
calculate the area of half the region and
double the result.
* Multiple Intersections: When the curves
intersect at multiple points, the region may
need to be divided into subregions. Each
subregion can then be analyzed separately,
and the total area is obtained by summing the
areas of these subregions.
* Polar Coordinates: For regions with circular
symmetry, using polar coordinates can often
lead to simpler integrals. By converting the
equations of the curves into polar form, we
can express the area as a double integral in
polar coordinates.
The ability to calculate areas bounded by
curves has far-reaching applications in various
fields:
* Physics:
 * Calculating work done by a variable force
* Determining the volume of solids of
revolution
* Analyzing the motion of objects under the
influence of varying forces
* Engineering:
* Designing structures and systems with
optimal dimensions and load-bearing
capacities
* Analyzing fluid flow and heat transfer in
various engineering systems
* Economics:
* Modeling consumer and producer surplus
in market analysis
* Studying the distribution of wealth and
income
* Statistics:
* Calculating probabilities and statistical
measures
* Analyzing data distributions and trends
By mastering the techniques of finding areas
bounded by curves, we can gain valuable
insights into a wide range of real-world
problems and make informed decisions based
on quantitative analysis.