Finding the Area of a Region Bounded by a Curve

[mathgirl89]

Find the exact area of the region bounded by the curve:
x = y^2 and the lines:
x = y - 1
y = -3
y = 3

so far, I have:

the integral from -3 to 3 OF (y^2)- (y-1)dy
(1/3) y^3 - (1/2) y^2 + y

top-bottom gives me:
9- 4.5 + 3 = 7.5 for (3)
and
-9 + 4.5 - 3 = -7.5

Subtract them and I get 0! I know this isn't correct. If anyone could tell me where I went wrong on this I'd greatly appreciate it!

Thank you in advance!

 

[galactus]

I believe this is what you're looking for. If I am interpreting correctly.

Breaking it upin to sections.

\(\displaystyle \L\\\int_{0}^{2}(x+1-\sqrt{x})dx+\int_{2}^{9}(3-\sqrt{x})dx+\int_{0}^{9}(3-\sqrt{x})dx\)

 

[daon]

Using Galactus' pretty picture, one could also do the following which might be easier on the powers+fractions etc.

\(\displaystyle \L \int _{-3}^1 \( y^2 \) dy + \int _1^3 \( y^2-(y-1) \) dy\)

 

[mathgirl89]

wow...

ok! Thank you both for your help! I'm still a little lost though. I ended up with 16 when I added the integrals together. I was told the answer was supposed to equal 24. I then thought to use the same techniques to find the area of the 'negative area' on the left side of th y-axis:

the integral from -3 to -1 of: y^2- (y-1) which gave me 14.666.... .When I add this to my previous total, I don't get 24.

Thank you again!

 

[pka]

Why are you all excluding the area on the left of the y-axis?
\(\displaystyle \L\int\limits_{ - 3}^3 {\left[ {y^2 - \left( {y - 1} \right)} \right]dy} = 24\)

 

[mathgirl89]

ok

that was...alot easier. turning it and just taking it all together...thank you so much!