MathWithNaziaa : Surface area and volume

Surface area and volume

Surface area and volume are calculated for any three-dimensional geometrical shape. The surface area of any given object is the area or region occupied by the surface of the object. Whereas volume is the amount of space available in an object.

In geometry, there are different shapes and sizes such as sphere, cube, cuboid, cone, cylinder, etc. Each shape has its surface area as well as volume. But in the case of two-dimensional figures like square, circle, rectangle, triangle, etc., we can measure only the area covered by these figures and there is no volume available. Now, let us see the formulas of surface areas and volumes for different 3d-shapes.

 

What is the Surface Area?

The space occupied by a two-dimensional flat surface is called the area. It is measured in square units. The area occupied by a three-dimensional object by its outer surface is called the surface area. It is also measured in square units.

Generally, Area can be of two types:

(i) Total Surface Area

(ii) Curved Surface Area/Lateral Surface Area

Total Surface Area

Total surface area refers to the area including the base(s) and the curved part. It is the total of the area covered by the surface of the object. If the shape has a curved surface and base, then the total area will be the sum of the two areas.

Curved Surface Area/Lateral Surface Area

Curved surface area refers to the area of only the curved part of the shape excluding its base(s). It is also referred to as lateral surface area for shapes such as a cylinder.

What is Volume?

The amount of space, measured in cubic units, that an object or substance occupies is called volume. Two-dimensional doesn’t have volume but has area only. For example, the Volume of the Circle cannot be found, though the Volume of the sphere can be. It is so because a sphere is a three-dimensional shape.

 

CONE:

This is a 3D geometric shape created by two line segments connecting a common point. This point is known as vertex or apex. This shape tapers from a smooth and flat base to the apex.

Here are the formulas of volume and surface area of cone- 

If the vertical height of the cone = h Radius of the base = r Slant height = l 

Then, 

volume = 1/3 π r2h

Total surface area = π r2+ πrl

Lateral area = πrl

Sphere:

It is a geometrical shape in 3D space that is similar to the outer surface of a planet or ball. Mathematically, it can be defined as a set of points that are situated at an equal distance from any given point. 

To solve mathematical problems related to the sphere, you need to understand the formulas of surface area and volume of spheres.

If the radius = r, 

Then,

 volume = 4/3 πr3

Surface area = 4π r2

Cylinder:

This is a solid curvilinear shape with the surface formed by equal and fixed-distant points from a given line. 

Now, the formula of volume and surface area of cylinder-

If the radius of a cylinder = r Height = h

The lateral surface area without the top and bottom surface = 2πrh

The total surface area including top and bottom surface = 2πr (r + h)

Volume = π r2h

 

Cube and Cuboid

A cube is formed with six square faces or sides, and each vertex is the meeting point of three sides. However, a cuboid also has six faces, but the cuboid faces can be any quadrilateral. Mostly, cuboid faces are rectangular

Following are the formulas of volume and surface area of a cube -

If the length of one side of a cube = a, Then, 

surface area = 6a3

Volume = x3

Now, let’s move to the surface area and volume formula of a cuboid-

If the height of a cuboid = h Length = l  Width = w

Then, 

surface area = 2 (hl + lw + wh)

Volume = hlw

With the help of these above formulas, you can easily solve volume and surface area problems of cone, cube, cuboid, sphere, cylinder. 

 

 

 

 

MathWithNaziaa : Rationalization of Denominator

Rationalizing the denominator

Algebra is a vast area of study in mathematics, which is associated with number theory, arithmetic, geometry, and its analysis. Algebra is related to the study of symbols and their manipulation with mathematical operations. The given article is a study of the denominator and numerator, rationalisation, and explaining why to rationalise the denominator.

Why rationalize the denominator?

While performing a basic operation we rationalise a denominator to get the calculation easier and obtain a rational number as a result. In the process of rationalisation, we exclude the square roots, cube roots, or any other radical expressions from the equation. Let’s see the method of rationalisation by an example.

Numerator and Denominator

The numerator is the top part of a fraction. It explains the number of counts of part of the object present in the given fraction. The term numerator is derived from the Latin word ”enumerate” which means ‘to count’. Some example, a numerator of the fraction 2/5 explains that the given object is divided into 5 equal parts and the fraction contains two of it.

The denominator is the bottom part of a fraction. It explains how many parts of a whole object are broken into. The term denominator itself is derived from the Latin word “nomen”. The denominator indicates the type of fraction described numerator. An example of a denominator that is a denominator of a fraction is, say, 5, then that indicates that the whole object is divided into 5 equal parts.

Rationalization

Rationalization is the process of attaining a rational number as

a result of multiplying a surd with a similar surd. The other surd that multiplies is term as the rationalizing factor (RF). The whole process of rationalization is carried out by moving the square roots or cube roots from the denominator to the numerator.

For example, to rationalize an expression x + √y

Rationalizing factor x – √y

now,

= (x + √y)(x – √y) = x² – (√y)²

= x² – y

While performing a basic operation we

rationalize a denominator to get the calculation easier and obtain a rational number as a result. In the process of rationalization, we exclude the square roots, cube roots, or any other radical expressions from the equation. Let’s see the method of rationalization by an example.

Rationalize the expression (√3 – 1)/(√3 + 1)

In the expression, rationalizing factor of denominator that would be √3 – 1. Multiplying and dividing the rationalizing factor of denominator, 

 = (√3 – 1)/(√3 + 1) × (√3 – 1)/ (√3 – 1)

= (√3 – 1)²/(√3)² – 1

By the formula (a – b)² = a² – 2ab + b²

= (√3)² – 2√3 × 1 + (1)²/ (3 – 1)

= 4 – 2√3/2

Taking 2 in common in numerator

= 2(2 – √3)/2

Cancelling common factors

 = 2 – √3

Sample Problems

Question 1: Rationalize 2√3/√3

Solution:

To rationalize the expression 2√3/√3 a rationalizing factor is needed which is √3.

Now,

= 2√3/√3 × √3/√3

= 2 × 3 /3

= 2

Question 2: Rationalize (2 + √3)/√3

Solution:

To rationalize the expression, 2 + √3/√3 we need a rationalizing factor which is √3.

= 2 + √3/√3 × √3/√3

= 2√3 + 3/3

Question 3: Rationalize 1/√x

Solution:

To rationalize the expression 1/√x, rationalizing factor is required which is √x.

= 1/(√x x) √x/√x.

= √x /x

Question 4: Rationalize the expression 32/5 – √7

Solution:

32/5 – √7 to rationalise this expression, rationalizing factor is needed which is 5 + √7

= 32/5 – √7 × 5 + √7/5 + √7

= 32(5 + √7)/(5 – √7)(5 + √7)

= (160 – 32√7)/(25 + 5√7 – 5√7 – 7)

= (160 – 32√7)/32

= 5 – √7

Question 5: Rationalize the denominator 5 – √3/2 + √3

Solution:

To rationalize the expression 5 – √3/2 + √3 a rationalizing factor is needed 2 – √3.

= 5 – √3/2 + √3 × 2 – √3/ 2 – √3

= (5 – √3)(2 – √3)/ (2)² – (√3)²

= 10 – 5√3 – 2√3 + 3/4 – 3

= 13 – 7√3/1

= 13 – 7√3

Question 6: Rationalize (√3 – 1)/(√3 + 1).

Solution:

To rationalize the expression (√3 – 1)/(√3 + 1) a rationalizing factor is needed √3 – 1

= √3 – 1/√3 + 1 × √3 – 1/√3 – 1

= (√3 – 1)²/(√3)² – (1)²

= (3 + 1 – 2√3)/(3 – 1)

= (4 – 2√3)/2

= 2 – √3

Examples of Rationalization of Denominator :

Summary of rationalization of denominators

Rationalizing the denominator means eliminating the radical expressions in the denominator so that we do not have square roots, cubic roots, or any other roots. The main idea in rationalizing denominators is to multiply the original fraction by an appropriate value so that after simplifying, the denominator no longer contains radicals.

When the denominator is a monomial, we can apply the fact that:

Therefore, we can multiply both the numerator and denominator by the radical expression. After simplifying, we will obtain an expression without radicals in the denominator.

On the other hand, if the denominator is a binomial, we have to use the conjugate of the binomial. The conjugate of a binomial is equal to the same binomial, but with the sign of the middle changed.

For example, suppose we have the binomial 

 in the denominator. The conjugate of this binomial is 

The product of the binomial and its conjugate is:

MathWithNaziaa : Integration by Partial Fractions and Integral Calculus Formulas

 

Integration by Partial Fractions

Integration by Partial Fractions: We know that a rational function is a ratio of two polynomials P(x)/Q(x), where Q(x) ≠ 0. Now, if the degree of P(x) is lesser than the degree of Q(x), then it is a proper fraction, else it is an improper fraction. Even if a fraction is improper, it can be reduced to a proper fraction by the long division process.

So, if P(x)/Q(x) is an improper fraction, then P(x)/Q(x) = T(x) + P1(x)/Q(x) … where T(x) is a polynomial and P1(x)/Q(x) is a proper rational fraction. We already know how to integrate polynomials and in this article, we will be learning about integration by partial fractions. Also, the rational functions that we will consider are those whose denominators can be factorised into linear and quadratic equations.

Different Forms Integration by Partial Fractions

Let’s say that we want to evaluate ∫ [P(x)/Q(x)] dx, where P(x)/Q(x) is a proper rational fraction. In such cases, it is possible to write the integrand as a sum of simpler rational functions by using partial fraction decomposition. Post this, integration can be carried out easily. The following image indicates some simple partial fractions which can be associated with various rational functions:

 

Please note that A, B, and C are real numbers and their values should be determined suitably.

Some Examples of Integration by Partial Fractions

Question 1:  Find ∫ dx / [(x + 1) (x + 2)]

Answer : The integrand is a proper rational function. Therefore, by using the form of partial fraction from the image above, we have:

1 / [(x + 1) (x + 2)] = A / (x + 1) + B / (x + 2) … (1)

Solving this equation, we get,

A (x + 2) + B (x + 1) = 1

Or, Ax + 2A + Bx + B = 1

x (A + B) + (2A + B) = 1

For LHS to be equal to RHS, we have

A + B = 0 and 2A + B = 1. On solving these two equations, we get

A = 1 and B = – 1.

Therefore, we have

1 / [(x + 1) (x + 2)] = 1 / (x + 1) – 1 / (x + 2)

Hence, ∫ dx / [(x + 1) (x + 2)] = ∫ dx / (x + 1) – ∫ dx / (x + 2)

= log |x + 1| – log |x + 2| + C

Note: Equation (1) is true for all permissible values of x. Some authors use the symbol ‘≡’ to indicate that the statement is an identity and use the symbol ‘=’ to indicate that the statement is an equation, i.e., to indicate that the statement is true only for certain values of x.

Question 2: Find ∫ [(x2 + 1) / (x2 – 5x + 6)] dx

Answer : In this case, the integrand is NOT a proper rational function. Hence we divide (x2 + 1) by (x2 – 5x + 6) and get,

(x2 + 1) / (x2 – 5x + 6) = 1 + (5x – 5) / (x2 – 5x + 6)

= 1 + (5x – 5) / (x – 2) (x – 3)

Now, let’s look at the second half of the above equation and let

(5x – 5) / (x – 2) (x – 3) = A / (x – 2) + B / (x – 3)

On solving it, we get

5x – 5 = A (x – 3) + B (x – 2) = Ax – 3A + Bx – 2B = x (A + B) – (3A + 2B)

Comparing the coefficients of the x term and constants, we get

A + B = 5 and 3A + 2B = 5. Further, on solving these two equations, we get

A = – 5 and B = 10.

Hence, we have

(x2 + 1) / (x2 – 5x + 6) = 1 – 5 / (x – 2) + 10 / (x – 3)

Therefore, ∫ [(x2 + 1) / (x2 – 5x + 6)] dx = ∫ dx – 5 ∫ 1 / (x – 2) + 10 ∫ 1 / (x – 3)

= x – 5log |x – 2| + 10log |x – 3| + C

Question 3: Find ∫ [(3x – 2) / (x + 1)2 (x + 3)] dx

Answer : If you look at the image above, then the denominator in this case is similar to example 4. Hence, we have

(3x – 2) / (x + 1)2 (x + 3) = A / (x + 1) + B / (x + 1)2 + C / (x + 3)

On solving it, we get

3x – 2 = A (x + 1) (x + 3) + B (x + 3) + C (x + 1)2

= A (x2 + 4x + 3) + B (x + 3) + C (x2 + 2x + 1) = Ax2 + 4Ax + 3A + Bx + 3B + Cx2 + 2Cx + C

= x2 (A + C) + x (4A + B + 2C) + (3A + 3B + C)

Comparing the coefficients of x2, x and the constant terms, we get

A + C = 0

4A + B + 2C = 3

3A + 3B + C = – 2

On solving these three equations, we get

A = 11/4, B = –5/2 and C = –11/4

Hence, we have

(3x – 2) / (x + 1)2 (x + 3) = 11 / 4(x + 1) – 5 / 2(x + 1)2 – 11 / 4(x + 3)

Therefore, ∫ [(3x – 2) / (x + 1)2 (x + 3)] dx = 11/4 ∫ dx / (x + 1) – 5/2 ∫ dx / (x + 1)2 – 11/4 ∫ dx / (x + 3)

= 11/4 log |x + 1| + 5 / 2(x + 1) – 11/4 log |x + 3| + C

= 11/4 log |(x + 1) / (x + 3)| + 5 / 2(x + 1) + C

Example 4: Find ∫ [(x2 + x + 1) / (x + 2) (x2 + 1)] dx

Solution: Now look at the example 5 in the image above. This example is similar to that example. Hence, we have

(x2 + x + 1) / (x + 2) (x2 + 1) = A / (x + 2) + (Bx + C) / (x2 + 1)

On solving this equation, we get

x2 + x + 1 = A (x2 + 1) + (Bx + C) (x + 2) = Ax2 + A + Bx2 + 2Bx + Cx + 2C

= x2 (A + B) + x (2B + C) + (A + 2C)

Comparing the coefficients of x2, x and the constant terms, we get

A + B = 1

2B + C = 1

A + 2C = 1

On solving these three equations, we get

A = 3/5, B = 2/5 and C = 1/5

Hence, we have

(x2 + x + 1) / (x + 2) (x2 + 1) = 3 / 5(x + 2) + [(2/5) x + 1/5] / (x2 + 1) = 3 / 5 (x + 2) + 1/5 [(2x + 1) / (x2 + 1)]

Therefore, ∫ [(x2 + x + 1) / (x + 2) (x2 + 1)] dx = 3/5 ∫ dx / (x + 2) + 1/5 ∫ [2x / (x2 + 1)] dx + 1/5 ∫ [1 / (x2 + 1)] dx

= 3/5 log |x + 2| + 1/5 log |x2 + 1| + 1/5 tan–1 x + C

More Solved Examples :

Question: Find ∫ [(3x – 1) / (x – 1) (x – 2) (x – 3)] dx

Solution: The integrand can be written as follows:

(3x – 1) / (x – 1) (x – 2) (x – 3) = A / (x – 1) + B / (x – 2) + C / (x – 3)

Multiplying both sides by {(x – 1) (x – 2) (x – 3)}, we get

3x – 1 = A (x – 2) (x – 3) + B (x – 1) (x – 3) + C (x – 1) (x – 2) … (a)

Now, let’s substitute x = 1 in equation (a). Hence, we have

3 – 1 = A (1 – 2) (1 – 3) + B (1 – 1) (1 – 3) + C (1 – 1) (1 – 2)

Or, 2 = A (–1) (–2) + B (0) (–2) + C (0) (–1) = 2A

Therefore, we get A = 1

Next, let’s substitute x = 2 in equation (a). Hence, we get

6 – 1 = {A (2 – 2) (2 – 3) + B (2 – 1) (2 – 3) + C (2 – 1) (2 – 2)}

Or, 5 = A (0) (–1) + B (1) (¬–1) + C (1) (0) = –B

Therefore, we get B = –5

Finally, let’s substitute x = 3 in equation (a). Hence, we get

9 – 1 = {A (3 – 2) (3 – 3) + B (3 – 1) (3 – 3) + C (3 – 1) (3 – 2)}

Or, 8 = A (1) (0) + B (2) (0) + C (2) (1) = 2C

Therefore, we get C = 4.

Hence, we now have

(3x – 1) / (x – 1) (x – 2) (x – 3) = 1 / (x – 1) – 5 / (x – 2) + 4 / (x – 3)

Therefore,

∫ [(3x – 1) / (x – 1) (x – 2) (x – 3)] dx = ∫ [1 / (x – 1)] dx – 5 ∫ [1 / (x – 2)] dx + 4 ∫ [1 / (x – 3)] dx

= log |x – 1| – 5log |x – 2| + 4 log |x – 3| + C

Solved Question :

Question 1: Why do we use partial fractions?

Answer: We decompose fractions into partial fractions like this as it makes specific integrals much simpler to do, and. We use it in the Laplace transform, which we meet later.

Question 2: How do you know when to use partial fractions in integration?

Answer: We can only do partial fractions if the degree of the numerator is severely less than the degree of the denominator. This is an important point. Thus, once you have understood that partial fractions can be done, you will need to factor the denominator as completely as possible

Question 3: How do you integrate fractions?

Answer: If someone asks you to integrate a fraction, you must try to multiply or divide the top and bottom of the fraction by a number. Occasionally it will be of help if you split a fraction up prior to making an attempt to integrate it. You can use the method of partial fractions for this.

Question 4: What is mean integration?

Answer: Integration happens when we bring distinct people or things together. For instance, the integration of students from all of the city’s colleges at the university and more. We have all come across the word differentiate that means to “set apart.” Thus, integrate is the opposite of this.

Integral Calculus Formulas

Integration is the reverse process of differentiation. So, we may call it as Inverse Differentiation. Integration is the process to find a function with its derivative. Basic integration formulas on different functions are very useful and important. This article deals with the concept of integral calculus formulas with concepts and examples. Integral calculus is the branch of mathematics dealing with the formulas for integration, and classification of integral formulas. The student will take benefits from this concrete article. Let us learn the concept and the integral calculus formulas.

Concept of the Integral Calculus

Integration is the algebraic method to find the integral for a function at any point on the graph. The integral comes from not only to determine the inverse process of taking the Derivative.

But also for solving the area problem as well. The integral of the function of x from range a to b will be the sum of the rectangles to the curve at each interval of change in x as the number of rectangles goes to infinity.

Given a function, f(x), is an anti-derivative of f(x), is any function F(x) such that

F ′ (x)=f(x)

If F(x) is any anti-derivative of f(x), then the most general anti-derivative of f(x) will be its indefinite integral.

The integral of a function f(x) with respect to variable x is:

∫f(x)dx

Also, integration is considered as almost an inverse to the operation of differentiation means that if,

then

∫g(x)dx=f(x)+C

 

The extra C called the constant of integration, which is really necessary. Definite integrals are the special kind of integration, where both endpoints are fixed. So, it always represents some bounded region, for computation.

Some Properties

1. ∫kf(x)dx=k∫f(x)dxwhere k is any number. So, we factor the multiplicative constants out of indefinite integrals.

1. ∫–f(x)dx=–∫f(x)dxThis is really the first property with k=−1k=−1 and so no proof of this property will be given.

1. ∫f(x)±g(x)dx=∫f(x)dx±∫g(x)dxIt means the integral of a sum or difference of functions is the sum or difference of their individual integrals. This is applicable and useful for many functions as we need.

Formula for Integral Calculus

 

Solved Examples for Integral Calculus Formulas