Integration: Area enclosed by the y-axis (1 Viewer)

[Makro]

7. Find the area bounded by the curve x = y² - 2y - 3 and the y axis.

I'm not sure what to do when you're not given the boundaries. I tried letting x = 0, to give the y intercepts, and I got y=3 and y=1 for them. End answer was wrong.

Correct answer: 10/2/3 units.

 

[shaon0]

I'm not sure what to do when you're not given the boundaries. I tried letting x = 0, to give the y intercepts, and I got y=3 and y=1 for them. End answer was wrong.

Correct answer: 10/2/3 units.

limits are: y=3, y=-1.
using those you get the answer.

 

[Makro]

I just wasn't thinking straight, heh. Also I'm not understanding where the abs. value signs go? Because I keep getting a negative. I did:

9 - 9 -9 - | -1/3 - 1 +3 |

Which gives - 10/2/3. From that I'd say it actually goes around the 9s. Though how are you supposed to know where to put the abs values? say if you can't draw the graph cos you're mathematically retarded or it's just a hard question. Is it just when you sub a limit in and the answer is negative? You whack it around that?

Thanks, and thanks in advance.

 

[Timothy.Siu]

well, y^2-2y-3 is a parabola and by integrating that with those boundaries it is negative because u are effectively finding the area bounded by y=x^2-2x-3 and the x-axis which is negative coz its under the x-axis

 

[Makro]

You mean "under" the y-axis? As in the negative side. If so I think I understand what you mean. Thanks.

Got another one that I failed at.

8. Find the area bounded by the curve x = -y² - 5y - 6 and the y axis.

My limits were 3 and 2, my answer was 24.8333 and the correct answer was 1/6.