# Evaluate the integral

#### Petrus

##### Well-known member
Hello MHB,
This is an exemple I do not understand.
if $$\displaystyle R={(x,y)|-1 \leq x \leq 1, -2 \leq y \leq 2}$$, evaluate the integral
$$\displaystyle \int\int_R \sqrt{1-x^2}dA$$ (It is suposed to be R at down for 'rectangle' if I understand correct.)
they solve it like this $$\displaystyle \int\int_R \sqrt{1-x^2}dA = \frac{1}{2} \pi(1)^2 * 4 = 2\pi$$
I don't understand how they do it and
The defination says:
"If $$\displaystyle f(x,y) \geq0$$, then the volume V of the solid that lies above the rectangle R and below the surface $$\displaystyle z=f(x,y)$$ is
$$\displaystyle V=\int\int_R f(x,y)dA$$"
What I can note is that $$\displaystyle \sqrt {1-x^2} \geq0$$ and $$\displaystyle z=\sqrt{1-x^2} <=> x^2+z^2=1$$ so then the volume lies above the rectangle and below $$\displaystyle z= \sqrt{1-x^2}$$ but I get confused with dA, integrate respect to area?

Regards,

Last edited:

#### ZaidAlyafey

##### Well-known member
MHB Math Helper
You are asked to solve the integral

$$\displaystyle \int^{2}_{-2} \, \int^{1}_{-1} \sqrt{1-x^2} \, dx \, dy$$

#### Petrus

##### Well-known member
You are asked to solve the integral

$$\displaystyle \int^{2}_{-2} \, \int^{1}_{-1} \sqrt{1-x^2} \, dx \, dy$$
is it that easy >.<

#### ZaidAlyafey

##### Well-known member
MHB Math Helper
Yes , you are asked to integrate over the rectangle [-1,1]x[-2,2]

#### Petrus

##### Well-known member
Yes , you are asked to integrate over the rectangle [-1,1]x[-2,2]
How will it work with integrate with dy? I dont have any y in my function so..?

#### ZaidAlyafey

##### Well-known member
MHB Math Helper
You can think of it as a nested integral so you start first by evaluating $$\displaystyle \int^1_{-1} \sqrt{1-x^2} \, dx$$

Then what ever the result you get assume to be f(y) then you evaluate

$$\displaystyle \int^{2}_{-2} f(y) \, dy$$

#### Petrus

##### Well-known member
You can think of it as a nested integral so you start first by evaluating $$\displaystyle \int^1_{-1} \sqrt{1-x^2} \, dx$$

Then what ever the result you get assume to be f(y) then you evaluate

$$\displaystyle \int^{2}_{-2} f(y) \, dy$$
Thanks Zaid! One last quest why can I do that? I dont think I ever read about it or I have just missed it. Thanks!

#### ZaidAlyafey

##### Well-known member
MHB Math Helper
This is what we call iterated integral .

- - - Updated - - -

Just as an exercise try to evaluate the integral of z=2 over the rectangle [0,2]x[0,2] and verify this is the volume of a cube.

#### Petrus

##### Well-known member
This is what we call iterated integral .

- - - Updated - - -

Just as an exercise try to evaluate the integral of z=2 over the rectangle [0,2]x[0,2] and verify this is the volume of a cube.
This is what makes this forum so good ! Thanks for helping me and giving me an exercise! I really like this thanks thanks! we got :
$$\displaystyle \int_0^2\int_0^22 dxdy$$ so the volume becomes $$\displaystyle V=8cm^3$$ (I just like to have 'unit'
A cube is like 6 square when our is 2cm long (Here is me draw it on paint but it looks like a rectangle but its a square) formel for volume of a cube is $$\displaystyle a^3$$ and our $$\displaystyle a=2$$ ( lenght) so the volume of the cube becomes $$\displaystyle 2^3=8cm^3$$

#### ZaidAlyafey

##### Well-known member
MHB Math Helper
Excellent , so we see that for positive z>0 this is actually the volume over the specified area on the xy-plane .