Area under curves

[klooless]

Hi there,

Question:
Find area bounded by the given curves by two methods (a) integrating with respect to x, and (b) integrating with respect to y.
4x+y^2 = 0 and y = 2x + 4

I'm having trouble finding the intersecting points for these two equations... help would be great!!

cheers

 

[Deleted member 4993]

Hi there,

Question:
Find area bounded by the given curves by two methods (a) integrating with respect to x, and (b) integrating with respect to y.
4x+y^2 = 0 and y = 2x + 4

I'm having trouble finding the intersecting points for these two equations... help would be great!!

cheers

4x + (2x+4)[sup:ceck9t8y]2[/sup:ceck9t8y] = 0

4x[sup:ceck9t8y]2[/sup:ceck9t8y] + 20x + 16 = 0

4(x + 1)(x + 4) = 0

Now continue....

 

[galactus]

If you graph them, it may help.

 

[BigGlenntheHeavy]

Integrating with respect to x:

\(\displaystyle \int_{-4}^{-1}(2x+4+2\sqrt-x)dx \ + \ \int_{-1}^{0}4\sqrt-xdx \ = \ 9.\)

integrating with respect to y:

\(\displaystyle \int_{-4}^{2}\frac{[-y^{2}-2y+8]}{4}dy \ = \ 9.\)

Note: y is better, only one integral.

 

[klooless]

Thanks for all the tips and help! Lot's of practice still to do, but you three helped tremendously!

Cheers!