Solve Riemannian Geometry (Omkar Nayak)

Introduction to solve Riemannian geometry:

Solve Riemannian geometry refers solving the Riemannian manifolds and smooth manifolds using the Riemannian metrics. Riemannian metrics are nothing but the tangent space with the curves inner product which varies in smooth manner from point to point. To solve the Riemannian geometry we will use the Riemannian sums and Riemannian integrals method. Basically Riemannian geometry refers the elliptic geometry. Here we are going to solve the area of the curve underneath. We will see some example problems for solve Riemannian geometry.

Solve Riemannian geometry - formulas:

If we have to solve the Riemannian geometry we have to use the Riemann sums and integrals method. Using this we have to find the area of the given curve on the graph underneath. In Riemannian geometry the Riemann sums and integrals used in definite integration operation. The Riemann integral is defined by taking the limit for the given Riemann sums. It is based on Jordan measure.

If we want to use the Riemannian sums the formula is

'S = sum_(i = 1)^nf(y_i)(x_i - x_(i - 1))' Here xi - 1= y i= x. Here the choice of y i is the arbitrary.

If the y i = xi - 1 is for all i values then it is called Left Riemann sum.

If the y i = xi then it is called right Riemann sum.

The average of the above two Riemannian is called Trapezoidal sum.

If the y i = (xi - xi - 1) / 2 then we can call this as middle Riemann sum.

If we want to use the Riemannian integrals the formula is

' int_a^bf(x)dx=lim_(maxDeltax->0)sum_(k=1)^nf(x^n)Deltax'

Examples for solve Riemannian geometry:

Examples 1 for solve Riemannian geometry:

Find the area of the given curve under y = x2 among the limits 0 and 3 using Riemannian sum.

Solution:

The area below the curve of x2 among the limits 0 and 3 may be computed procedurally using the Riemann's Sum method. The interval 0 and 3 is divided into n number of sub intervals. Each sub interval gives the width of the 3/n. These are called width of the Riemann's rectangles. The sequence of all x coordinates can be defined as X1, X2 . . . , X n. Then the heights of the Riemann Rectangle boxes can be defined by the following (X1)2, (X2)2 . . . , (X n) 2. This is an important fact where Xi ='(3i) / n' .

The area of a single box will be (3 / n)(xi) 2

S =' (3 / n) xx (3 / n)^2 + . . . . + (3 / n) xx ((3i) / n) ^2+ . . . +(3 / n) xx (3)^2'

S = '27 / n^3 (1 + . . . + i^2 + . . . . + n^2)'

S = '27 / n^3((n(n + 1)(2n + 1)) / 6)'

S = '27 / n^3 ((2n^3 + 3n^2 + n) / 6)'

S =' 27 / 3 + 27 / (2n) + 27 / (6n^2 )'

S =' lim_(n-gtoo)( 27 / 3 + 27 / (2n) + 27 / (6n^2 ))'

S = '27 / 3' = 9

Examples 2 for solve Riemannian geometry:

Find the area of the curve under y = x3 among the limits 0 and 3 using Riemannian integral.

Solution:

In Riemann integrals help we can calculate the area above for the interval 0 and 3. Here

Riemann integral =' int_0^3(x^3)= (x^4 / 4)'

Now we have to take the limit is 0 and 3

If we applying the limit from 0 to 3 we get

= '3^4 / 5 - 0^4 / 4 = 81 / 4'

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