Thread: Finding the Area Between 2 Curves

1. Finding the Area Between 2 Curves

I have two problems I need help on:

1. Find the area of the region lying to the right of x=(y^2)-5 and the left of x=3-(y^2).

I determined that the limits of integration would be [-5, 3] and then I tried to integrate ((y^2)-5)-(3-(y^2)), but I keep getting a REALLY large unrealistic number,

2. Calculate the area enclosed by x=9-y^2 and x=5 in two ways: as an integral along the y-axis and as an integral along the x-axis.

I'm not sure how to start this one.

Any help will be so greatly appreciated.

2. Originally Posted by KDonicki
I have two problems I need help on:

1. Find the area of the region lying to the right of x=(y^2)-5 and the left of x=3-(y^2).

I determined that the limits of integration would be [-5, 3] and then I tried to integrate ((y^2)-5)-(3-(y^2)), but I keep getting a REALLY large unrealistic number,
First have you sketched the curves?

Code:
EuMathT code to sketch curves

>y=-6:0.1:6;
>
>x1=y^2-5;
>x2=3-y^2;
>
>
>xplot(x1,y);
>hold on;plot(x2,y);
>hold off
>

Sketch attached.

So you see that you need to integrate:

$\int_{y=-2}^2 (y^2-5)-(3-y^2) dy$

RonL

3. Originally Posted by KDonicki

2. Calculate the area enclosed by x=9-y^2 and x=5 in two ways: as an integral along the y-axis and as an integral along the x-axis.

I'm not sure how to start this one.
Again, sketch the curves, and then see what you can do.

RonL

4. for the second problem I used the [-2, 2] as my limits of integration and then I integrated this:

$
\int_{y=-2}^2 (9-y^2) dy
$

since (9-y^2) is the "upper" function (since its more on the right). However I keep getting, 92/3, but the back of my book says 32/3. What am I doing wrong?

5. Why are you doing 9 - y^2?

$
\int_{-2}^2 (9-y^2) - 5 dy = \int_{-2}^2 (4-y^2) dy
$