area under curves

[Guest]

I am stumped... y = 4 - x^2 and the x-axis, from y = 0 to y=3 .. wondering if i have to just rearrange the y= to x= and solve normally or if there is another way? Thanks!

 

[galactus]

You could solve for x and use the y limits you're given:

Since you'll have a square root, you can use \(\displaystyle 2\int_{0}^{3}{sqrt{4-y}}dy\)

If you want to use x, sub in the y values given into \(\displaystyle y=4-x^{2}\)

So, x=-1, 1, -2, 2

\(\displaystyle \int_{-2}^{2}{4-x^{2}}dx-\int_{-1}^{1}{(4-x^{2}-3)}dx\)

 

[soroban]

Hello, abrar!

\(\displaystyle y\:=\:4\,-\,x^2\) and the x-axis, from \(\displaystyle y = 0\) to \(\displaystyle y=3\)

Wondering if i have to just rearrange the y= to x= and solve normally \(\displaystyle \;\) . . . this works
or if there is another way? \(\displaystyle \;\) . . . yes

      |      ***
      |   *---+---*
      | *:|:::|:::|:*
      |*::|:::|:::|::*
      |:::|:::|:::|:::
     -*---+---+---+---*-
     -2  -1   |   1   2

We have: \(\displaystyle \,x\:=\:\sqrt{4\,-\,y}\)

The area is: \(\displaystyle \L\:A\;= \;2 \int^{\;\;\;3}_0(4\,-\,y)^{\frac{1}{2}}dy\)


Another way:

Note that there is a 2-by-3 rectangle in the middle (area 6).
\(\displaystyle \;\;\)We need the area of the regions at the far left and far right.

One of them is: \(\displaystyle \L\:\int^{\;\;\;2}_1(4\,-\,x^2)\,dx\)


Either way, the answer is: \(\displaystyle \L\,\frac{28}{3}\)

 

[Guest]

thank you so much, that really helped!!