# Finding the Area Between 2 Curves

• Feb 1st 2008, 10:06 PM
KDonicki
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.
• Feb 1st 2008, 10:31 PM
CaptainBlack
Quote:

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:

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

RonL
• Feb 1st 2008, 10:47 PM
CaptainBlack
Quote:

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
• Feb 2nd 2008, 10:11 AM
KDonicki
for the second problem I used the [-2, 2] as my limits of integration and then I integrated this:

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

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? :(
• Feb 2nd 2008, 10:33 AM
TrevorP
Why are you doing 9 - y^2?

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