Question

area between 2 curves. please help

Answer 1

I have to find the area between \[x=y^2+1\]\[and\]\[x=9-y^2\]

Answer 2

this is the picture I got

Answer 3

but I have 1 question

Answer 4

will the area below the x-axis cancel the area above the x-axis ?

Answer 5

Graph them both and see what is the area between the curves that overlaps.

Answer 6

so, the area below the x-axis has to be canceling the area above the x-axis in any case where I am finding area between 2 curves?

Answer 7

if so the answer is 0, and if not, then...

Answer 8

if not, I will re-write the functions in terms of function y(x),
integrate the first from 1 to 5,
and the second from 5 to 9, and add the areas
(including only positive root) and then mutiply the sum times 2

Answer 9

yes, I got the desmos picture

Answer 10

\[y=\sqrt{x-1}\]\[y=\sqrt{9-x}\]\[A(x)=2\left(\int\limits_{1}^{5} (\sqrt{x-1})dx~+~\int\limits_{5}^{9}(\sqrt{9-x})~dx\right)\]

Answer 11

this is if the area under the x-axis does NOT cancel with the area above the x-axis.

Answer 12

and the part I don't know, is whether the area below and above x-axis cancel each other or not,

Answer 13

What value did you get after solving the definite integral?

Answer 14

I am too lazy now, I will put it into wolfram

Answer 15

I will see if the value is zero or not.

Answer 16

http://www.wolframalpha.com/input/?i=2%28+integral+from+1+to+5+%E2%88%9A%28x-1%29+%2B+integral+from+5+to+9+%E2%88%9A%289-x%29+%29

64/3

Answer 17

well, I did the area above the x-axis times two

Answer 18

so that is positive.

Answer 19

and again, this is where my problem is.
Is it correct to just multiply the area bounded by the two curves and the x-axis times 2, or do area below and above the x-axis cancel.

do you know ?

Answer 20

How come you multiplied the integrals by 2?

Answer 21

lol sorry, i'm just a little lost here. Just covered this topic yesterday xD

Answer 22

the integrals inside the parenthesis are the area between the 2 curves that is above the x-axis. Correct ?

Answer 23

then, there is the exact mirror/reflection of this shape/area below the x-axis.

Answer 24

or lost?

Answer 25

https://www.desmos.com/calculator/zsteb4lgbv
this is I re-wrote the functions in terms of y=f(x)

BUT without including the +- by the square roots, which therefore gives me just the area above the x-axis.

Answer 26

and then, after finding all the area between the curves which is above the x-axis, I multiply the result times 2.
Why do multiply times 2, because there are 2 exactly same regions.
1) below x-axis
2) above x-axis

Answer 27

this is all, IFF areas above and below x-axis don't cancel each other.
I know how to calculate, but now what to calculate... what a pain -:(

Answer 28

@perl

Answer 29

tagged:)

Answer 30

X=y^2+1
x=9-x^2
\[dA=(X-x)dy\]
\(dA=(y^2+1-9-x^2)dy\)
\[\int\limits_{-2}^{2}(y^2+1-9-y^2)dy\]
I chose to do with the respect to y, you can also do with the respect to x