Category Archives: Integration

Useful trigonometric formulas for finding areas of circles

If you’re trying to find the area of a circle using integration methods, then these trigonometric formulas are going to be very useful:

First formulas:

\sin ^{ 2 }{ \theta +\cos ^{ 2 }{ \theta =1 } } \\ \\ \therefore \quad { r }^{ 2 }\sin ^{ 2 }{ \theta +{ r }^{ 2 }\cos ^{ 2 }{ \theta ={ r }^{ 2 } } } \\ \\ \therefore \quad { r }^{ 2 }\sin ^{ 2 }{ \theta ={ r }^{ 2 }-{ r }^{ 2 }\cos ^{ 2 }{ \theta } } \\ \\ \therefore \quad { r }^{ 2 }\sin ^{ 2 }{ \theta ={ r }^{ 2 }\left( 1-\cos ^{ 2 }{ \theta } \right) }

Second formulas:

\sin ^{ 2 }{ \theta +\cos ^{ 2 }{ \theta =1 } } \\ \\ \therefore \quad { r }^{ 2 }\sin ^{ 2 }{ \theta +{ r }^{ 2 }\cos ^{ 2 }{ \theta ={ r }^{ 2 } } } \\ \\ \therefore \quad { r }^{ 2 }\cos ^{ 2 }{ \theta ={ r }^{ 2 }-{ r }^{ 2 }\sin ^{ 2 }{ \theta } } \\ \\ \therefore \quad { r }^{ 2 }\cos ^{ 2 }{ \theta ={ r }^{ 2 }\left( 1-\sin ^{ 2 }{ \theta } \right) }

These formulas are to be used when you have to transform the expression:

{ y }=\sqrt { { r }^{ 2 }-{ x }^{ 2 } }

You can either make:

x=r\sin { \theta }

Or…

x=r\cos { \theta }

The choice is yours. ūüôā

Derive the formula to find areas underneath curves

In this post I’ll be revealing how you can derive the formula which can be used to find areas underneath curves, from absolute scratch. Now, just below, what you will find is the diagram that will help us produce this formula…

area_underneath_curveIn this diagram what you will discover is that:

*Please read the following contents carefully

  • A length a exists, which starts at the origin O and ends at a;
  • A length¬†x¬†exists, which starts at the origin O and ends at x;
  • A length¬†x+?x exists, which starts at the origin O and ends at x+?x;
  • A length¬†?x exists, which starts at x and ends at x+?x;
  • A height y exists, which starts at the origin O and ends at y;
  • A height y+?y exists, which starts at the origin O and ends at y+?y;
  • A height¬†?y exists, which starts at y and ends at y+?y;
  • There is a curve called y=f(x);
  • There is an area underneath the curve called A which commences at a and ends at x;
  • There is an area underneath the curve called¬†?A which¬†commences at x and ends at x+?x¬†(Note: If you extend the distance from a to x what you get is a larger area, and the change in area can be measured. This change or difference is called ?A);
  • There is a rectangle that exists called QRUT. It has an area which is y?x;
  • There is a rectangle that exists called PRUS. It has an area which is (y+?y)?x;
  • ?A has an area larger than that of the rectangle QRUT, but smaller than that of the rectangle PRUS.

Producing the formula with the information we’ve discovered…

Ok, so we want to produce the formula which will help us find areas underneath curves from absolute scratch. At our disposal we have a helpful diagram (which we’ve looked at and analysed carefully) and we’ve been able to discover a few facts about it. I think we can now get to work…

Let’s start off by saying that:

Area QRUT < ?A < Area PRUS

Which is something we already discovered.

If this is the case, we can say that:

y?x < ?A < (y+?y)?x

Now, check out what happens when we divide all the elements of this expression by ?x:

What we end up with is…

daum_equation_1476040938725

 

 

Alright, now you may be saying to yourself, why do I need to know this? Well, it turns out that:

daum_equation_1476041141723

 

 

This is because as ?x approaches 0,  ?y approaches 0 leaving (?A)/(?x) sandwiched between y and y+0.000000000000000001 which is virtually y.

And, also…

daum_equation_1476041503530

 

 

As a consequence, this ultimately means that:

y=\frac { dA }{ dx }

This is incredibly significant, because if we then integrate both sides of this equation, we get:

\int { ydx=\int { \frac { dA }{ dx } } } dx\quad \Rightarrow \quad A=\int { ydx }

And…

A=\int { ydx } =F\left( x \right) +C

Now, this equation can actually be used to find the area A underneath the curve from a to x. What we’re basically saying is that this area is equal to some function of x plus a constant. This ‘some function of x’ occurs when we integrate y which is a function of x.


Finalising the formula…

Alright so we’ve managed to latch on to something incredibly significant… We’ve got an important equation:

A=\int { ydx } =F\left( x \right) +C

However, it is not complete. We need to know what the constant C is. So…

If we say that at x=a the area A underneath the curve¬†is 0, watch what happens… Look at what we get…

O=F\left( a \right) +C

Which means that:

C=-F\left( a \right) 

Hence, we can conclude that:

A=\int { ydx=F\left( x \right)  } -F\left( a \right) 

And this formula can be transformed into something more fancy if we are measuring an area underneath a curve from x=a to x=b

This is probably the formula you’re most familiar with…

A=\int _{ a }^{ b }{ ydx=F\left( b \right)  } -F\left( a \right) ={ \left[ F\left( x \right)  \right]  }_{ a }^{ b }

Which is the formula which can be used to find areas underneath curves.


If you are still confused and would like to go through this proof once again, please watch my video below…

You can also leave your comments below.


Related:

Trapezium Rule Formula – Derivation

Tricky Edexcel C4 Examination Question Solved (January 2008) Paper

Question (8):

Liquid is pouring into a large vertical circular cylinder at a constant rate of 1600 cm^3 per second and is leaking out of a hole in the base, at a rate proportional to the square root of the height of the liquid already in the cylinder. The area of the circular cross section of the cylinder is 4000 cm^2.

a) Show that at time t seconds, the height h cm of liquid in the cylinder satisfies the differential equation:

\frac { dh }{ dt } =0.4-k\sqrt { h } , where k is a positive constant.

Solution:

Well firstly let’s draw a little diagram to depict what is actually happening:

cylinder problem

 

As you can see, dV/dt is the rate at which the volume of liquid in the cylinder is changing over time. You have to remember that liquid is being poured into  the large vertical cylinder at a constant rate of 1600 cm^3 per second. What you also have to note is that the vertical cylinder is losing liquid at a rate proportional to the square root of the height already in the cylinder. Therefore we have \frac { dV }{ dt } =1600-C\sqrt { h } .

Now the area of the circular cross section should be: A=\pi { r }^{ 2 }. However, we are told that A=4000. Knowing that the Volume of the cylinder is V=\pi { r }^{ 2 }h, we can transform the volume formula to V=4000h.

Ok, so how can we get dh/dt? This is the formula we can use to derive it:

\frac { dh }{ dt } =\frac { dh }{ dV } \cdot \frac { dV }{ dt }

We know that:

V=4000h,\\ \\ \therefore \quad \frac { dV }{ dh } =4000,\quad and\quad as\quad \frac { dh }{ dV } =\frac { 1 }{ \frac { dV }{ dh } } ,\\ \frac { dh }{ dV } =\frac { 1 }{ 4000 } .

We also know that:

\frac { dV }{ dt } =1600-C\sqrt { h }

So what we get is:

\frac { dh }{ dt } =\frac { 1 }{ 4000 } \cdot \left( 1600-C\sqrt { h } \right) \\ \\ =\frac { 1600 }{ 4000 } -\frac { C }{ 4000 } \sqrt { h } \\ \\ =0.4-k\sqrt { h } ,\\ \\ \therefore \quad k=\frac { C }{ 4000 } .

—————————————————————————

b) When h=25, water is leaking out of the hole at 400 cm^3 per second. Show that k=0.02.

Solution:

Ok, with this information we can say that:

\frac { dV }{ dt } =1600-400

Moving on, we can determine that:

\frac { dV }{ dt } =1600-400,\quad but\quad \frac { dV }{ dt } =1600-C\sqrt { h } .\\ \\ Since,\quad h=25,\quad and\quad C\sqrt { h } =400,\\ \\ 5C=400,\quad \therefore \quad C=80.\\ \\ But\quad k=\frac { C }{ 4000 } =\frac { 80 }{ 4000 } ,\\ \\ \therefore \quad k=0.02.\\ \\

—————————————————————

c) Separate the variables of the differential equation:

\frac { dh }{ dt } =0.4-0.02\sqrt { h } \\ \\,

to show that the time taken to fill the cylinder from empty to a height of 100cm is given by:

\int _{ 0 }^{ 100 }{ \frac { 50 }{ 20-\sqrt { h } } } dh\\ \\

Solution:

\frac { dh }{ dt } =\frac { 2 }{ 5 } -\frac { \sqrt { h } }{ 50 } =\frac { 20 }{ 50 } -\frac { \sqrt { h } }{ 50 } \\ \\ =\frac { 20-\sqrt { h } }{ 50 } \\ \\ \frac { dh }{ dt } =\frac { 20-\sqrt { h } }{ 50 } \\ \\ dh=\frac { 20-\sqrt { h } }{ 50 } dt\\ \\ 50dh=20-\sqrt { h } dt\\ \\ \frac { 50 }{ 20-\sqrt { h } } dh=1dt\\ \\ \int { 1dt=\int { \frac { 50 }{ 20-\sqrt { h } } } } dh\\ \\ \therefore \quad t=\int _{ 0 }^{ 100 }{ \frac { 50 }{ 20-\sqrt { h } } } dh\\ \\

———————————————————————

d) Using the substitution:

h={ \left( 20-x \right) }^{ 2 }\\ \\ , or otherwise, find the exact value of:

\int _{ 0 }^{ 100 }{ \frac { 50 }{ 20-\sqrt { h } } } dh\\ \\

Solution:

If\quad h={ \left( 20-x \right) }^{ 2 },\quad \\ \\ { h }^{ \frac { 1 }{ 2 } }=\sqrt { h } ={ \left[ { \left( 20-x \right) }^{ 2 } \right] }^{ \frac { 1 }{ 2 } }=\left( 20-x \right) .\\ \\ If\quad h={ \left( 20-x \right) }^{ 2 }={ p }^{ 2 },\quad \frac { dh }{ dp } =2p,\quad p=20-x\quad \therefore \quad \frac { dp }{ dx } =-1\\ \\ So\quad \frac { dh }{ dx } =-2\left( 20-x \right) ,\quad \therefore \quad dh=-2\left( 20-x \right) dx\\ \\ When\quad h=100,\quad { \left( 20-x \right) }^{ 2 }=100,\quad \therefore \quad x=10.\\ When\quad h=0,\quad { \left( 20-x \right) }^{ 2 }=0,\quad \therefore \quad x=20.\

Therefore, we need to integrate:

\int _{ 20 }^{ 10 }{ \frac { 50 }{ 20-\left( 20-x \right) } \cdot \frac { -2\left( 20-x \right) }{ 1 } } dx=\int _{ 20 }^{ 10 }{ \frac { -100\left( 20-x \right) }{ x } } dx\\ \\ =\int _{ 20 }^{ 10 }{ \frac { -2000+100x }{ x } } dx=\int _{ 20 }^{ 10 }{ -\frac { 2000 }{ x } } +\frac { 100x }{ x } dx\\ \\ =\int _{ 20 }^{ 10 }{ -\frac { 2000 }{ x } } +100dx.\\ \\ But\quad as\quad -\int _{ a }^{ b }{ f\left( x \right) } dx=\int _{ b }^{ a }{ f\left( x \right) dx } ,\\ \\ \int _{ 20 }^{ 10 }{ -\frac { 2000 }{ x } } +100dx=\int _{ 10 }^{ 20 }{ \frac { 2000 }{ x } } -100dx\\ \\ ={ \left[ 2000\ln { x } -100x \right] }_{ 10 }^{ 20 }\\ \\ =2000\ln { 20 } -2000-\left( 2000\ln { 10-1000 } \right) \\ \\ =2000\ln { 20 } -2000-2000\ln { 10 } +1000\\ \\ =2000\left( \ln { 20-\ln { 10 } } \right) -2000+1000\\ \\ =2000\ln { 2-1000 }

——————————————————————–

e) Hence find the time taken to fill the cylinder from empty to a height of 100cm, giving your answer in minutes and seconds to the nearest second.

Solution:

Remember that:

t=\int _{ 0 }^{ 100 }{ \frac { 50 }{ 20-\sqrt { h } } } dh.

As this is the case,

t=2000\ln { 2 } -1000=386\quad seconds=6\quad minutes\quad and\quad 26\quad seconds.