Question

Who can help me find area between curves? 7x+y^2=8 x=y i dont know how to approach this one.

Answer 1

x=1/64 y=1/8 idk if that wat u was lookn for

Answer 2

i need to find shared area between these 2 curves

Answer 3

Solve both curves for x, the reason for solving for x is because you don't have to deal with the +/- ambiguity that would come from solving for y. So you have:
x=(1/7)(8-y^2)
Now, plug in a y value to see which is on the right. In this case, x=y is on the LEFT so you have:
(1/7)(8-y^2)=y
y^2+7y-8=0
(y+8)(y-1)=0
So your limits of integration are -8 to 1.
Your integral then looks like this:
\[\int\limits_{-8}^{1} \frac{1}{7}(8-y^2)-ydy=\frac{1}{7}\int\limits_{-8}^{1}8-y^2-7ydy=\frac{1}{7}[8y-\frac{y^3}{3}-\frac{7y^2}{2}]_{-8}^{1}\]

Answer 4

Which gives you 243/14.

Answer 5

wait a sec, i'll check it ...

Answer 6

I just ran it through wolfram.

Answer 7

thanks, i was confused on the part, where you need to find limits of integration..

Answer 8

Okay, all I did was set the equations(after solving them for x) equal.
So you have x=y; x=(1/7)(8-y^2)
Set them equal:
y=(1/7)(8-y^2)
Multiply by 7
7y=8-y^2
Add the y^2, subtract the 8
y^2+7y-8=0
Factor:
(y+8)(y-1)=0 =>y=-8,1

Answer 9

Is that all you need?

Answer 10

yea, thanks!

Answer 11

No problem :P