Question

I need to find the area of: \[D_5=\left\{ (x,y)\in \mathbb{R} : 0\le x \le x^2+y^2 \le 1\right\}\] Would it make sense to use polar coordinates? Like \[x=r*cos(a)\]and \[y=r*sin(a)\]?

Answer 1

u mean area of a unit circle?

Answer 2

Not sure, but since there is \[x^2+y^2\] It could be some kind of circle

Answer 3

Usually I would just use Fobinis theorem, but I am having trouble because I have \[ 0\le x \le x^2+y^2 \le 1\] Instead of something where x and y are seperate

Answer 4

i dont understand what sort of region we are looking at...

but generally presence of \(x^2+y^2\) terms suggests that a switch to polar is convenient

Answer 5

This is the region:\[D_5=\left\{ (x,y)\in \mathbb{R}^2 : 0\le x \le x^2+y^2 \le 1\right\}\]

Answer 6

Yea, thats what I though aswell.

Answer 7

So that would give: \[D_5=\left\{ (x,y)\in \mathbb{R} : 0\le r*cos(a) \le r^2 \le 1\right\}\] Now how would I make the double integral from this? - Using Fubinis theorem

Answer 8

How do you know it is 0.5?

Answer 9

\(x \le x^2+y^2\)

rearrange above inequality and put it in standard form

Answer 10

Soo, to see the circles, you split it up like this: \[D_5=\left\{ (x,y)\in \mathbb{R} : 0\le x \le x^2+y^2 \le 1\right\}\]Becomes
\[D_5=\left\{ (x,y)\in \mathbb{R} : 0\le x \le x^2+y^2,x \le x^2+y^2 \le 1\right\}\]

Answer 11

I guess yeah. Looks we need to include the region in IVth quadrant too

Answer 12

because \(0\le x\) gives you both I and IV quadrants

Answer 13

Yea, so that would give an area of \[A(D_5)=\frac{\pi}{2}\]

Answer 14

shaded region = \(\frac{1}{2}[\pi*1^2 - \pi*(\frac{1}{2})^2] =\frac{3\pi}{8} \)

right ?

Answer 15

Yea, right \[A=\pi*r^2\],this cannot be the answer though. (It is an exam from last year, that I am practicing. And you are supposed to insert the missing number from 1-99) And I ahve \[A(D_5)=\frac{\pi}{insert}\]

Answer 16

that is so wrong, wait

the center of smaller circle is not at origin

Answer 17

i mean, our earlier work is wrong

Answer 18

lets start over

Answer 19

Aight, you dont think that I should be using Fubinis theorem here? First converting it into polar coordinates, then setting up the double integral?

Answer 20

before thinking of setting up an integral to find the area of region, we need to figure out exactly what area we need to find