Monday

July 28, 2014

July 28, 2014

Posted by **Jenna** on Thursday, September 10, 2009 at 7:39pm.

Find the area of the region enclosed by the lines and curves:

x-(y^2)=0 and x+2(y^2)=3

Here is my work:

x= (y^2) and x= 3-2(y^2)

(y^2)= 3-2(y^2)

3(y^2)-3=0

3((y^2)-1)=0

3(y+1)(y-1)=0

y=-1, 1

That's following what the book showed my to do. However, the book used those y values as the upper and lower bounds of the integral, but by looking at the graph that doesn't make any sense. If I used those as the upper and lower bounds, the answer I got was 4 units^2. How do I do this problem?

- Calculus -
**Jenna**, Thursday, September 10, 2009 at 7:58pmI still keep getting 4. At this step:

area= int (3-2y^2-y^2) dy

= int (3(y^2-1) dy

shouldn't it be int(3(1-y^2)?

I did this:

3int(dy)-3int(y^2)dy= 3y- (3(y^3)/3) = 3y-(y^3) from 1 to -1... filling those in, I get 4. Am I still making a mistake?

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