area between to curves

[calchere]

Decide whether to integrate with respect to x or y. Then find the area of the region.

x=2y^2 , x+y=1

First off, I'm not sure whether to integrate with respect ot x or y.
I tried y and got:

y=sqrt(x/2) , y=1-x

But there is only one intersection point. x=1/2
I'm not sure what to do.
The answer is 9/8 .

 

[pka]

Here is the graph.

 

[calchere]

Ok so in terms of y it wont give me the negative part of the graph, so i tried in terms of x this time.
x=2y^2 , x= 1-y
y=.5 , y=-1

so i did x1 - x2
int[-1,.5] 1-y-2y^2

y-(y^2/2)-6(y^2)/3 from -1 to .5

[.5] - [-.125] = 3/8 but the answer is 9/8

 

[pka]

Yes that is correct.

 

[skeeter]

Ok so in terms of y it wont give me the negative part of the graph, so i tried in terms of x this time.
x=2y^2 , x= 1-y
y=.5 , y=-1

so i did x1 - x2
int[-1,.5] 1-y-2y^2

y-(y^2/2)-6(y^2)/3 from -1 to .5

incorrect antiderivative ... should be 2y³/3

[.5] - [-.125] = 3/8 but the answer is 9/8

 

[calchere]

Ah, i forgot to add one to the power.
Thank You both.

 

[galactus]

You could also try the x version. It's a little tougher than the y but good practice.



\(\displaystyle \L\\2\int_{0}^{\frac{1}{2}}\sqrt{\frac{x}{2}}dx+\int_{\frac{1}{2}}^{2}(1-x+\frac{\sqrt{x}}{2})dx\)