Integral Calculus

Question 1

Introduction

Answer 1

a single integral can be thought of as the area under the curve between two points approximated with rectangles.

If we take the limit of the number of rectangles to infinity, we get our 1D integral
integral(a-b) (fx) dx

Question 2

Double integrals

Answer 2

z = f(x,y)
When we have a function of two variables, we can sum up cuboids with base area dA to find the volume under the 2D curve.

By taking the limit of the number of cuboids to infinity, we have a double integral.
volume = double integral ((limits R at bottom) ( f(x,y) dA))

where R is the 2D region of integration.

Question 3

Examples of double integrals

Answer 3

Volume = double integral ((limits:R at bottom)(f(x,y))dA)

Area = double integral ((limits: R at bottom)(1)dA)

Second moment of area e.g
Ixx = double integral ((limits: R at bottom)(y^2)dA)

Centre of mass x = 1/M double integral( (xp(x,y)dA)

Question 4

Iterated integral

Answer 4

Can be expressed as a replacement for double integral. This allows us to evaluate the integral by doing normal integration twice.

Question 5

Extra notes

Answer 5

We find the volume under the surface f(x,y) by splitting the calculation into two parts:

Evaluate the inner integral to find the equation for the area of a 2D ‘slice’ of the volume
while holding the other variable constant.
This method is also known as partial integration.

The volume is found by outer integral summing these areas together for an arbitrarily large number of slices.

Question 6

Non rectangular regions

Answer 6

the integration limits contain x and y

we can follow the following steps:

1. Sketch the integration region,R , on the x-y axis.
2. Where limits are not given, draw a strip in the direction of the inner integral(while holding the outer integral constant) to find appropriate limits.

Question 7

Change of order

Answer 7

1. Sketch the integration region R, on the x-y asxis.
2. Swap the order of dx and dy for the inner and outer integrals
3. Use horizontal /vertical strips to determine new inner integral limits.
4. Use inner integral strips and the region R to determine new outer integral limits.

A few things to remember:
If the inner integral is dx, draw horizontal strips on R. The lower/upper limits are functions of left / right ends of the horizontal strips.

If the inner integral is dy, draw vertical strips on R. The lower/ upper limits are functions of bottom / top end of the vertical strips

Outer integral limits should contain CONSTANT VALUES ONLY.
Limits are equal to the range of values along the variable of the outer integral for the region R.

Question 8

Change of Variable

Answer 8

General procedure

1. Identify an appropriate coordinate system , If required.
2. Find the Jacobian for the proposed coordinate system
3. Substitute the new variables into the integral
4. Draw the integration region.
5. Draw strips in the direction of the inner integral to find the inner limits.
6. Add up strips across the region R to find the outer limits.

Question 9

Jacobian

Answer 9

j = r in polar coordinates
j = detriment of partial derivative of (dx/du, dx/dv,dy,du,dy,dv)