# Thread: Area between two curves problem

1. ## Area between two curves problem

Provided:

$\displaystyle y = 5 - x$

$\displaystyle x = (1/6)(y-5)^2$

How do I know to solve in terms of y or x?

So far I've tried it in terms of y and found the points of intersection to be -1, 5.. I tried evaluating the integral of $\displaystyle \int(5-y)-(1/6(y-5)^2)$, with the limits -1 to 5, but was marked incorrect... Any suggestions? Should it be in terms of x ?

Thanks,

John

2. You're integrating with respect to y, you need to find the limits along the y axis. You're using the limits along the x-axis. Try integrating from 0 to 6.

3. Originally Posted by JohnM25
Provided:

$\displaystyle y = 5 - x$

$\displaystyle x = (1/6)(y-5)^2$

How do I know to solve in terms of y or x?

So far I've tried it in terms of y and found the points of intersection to be -1, 5.. I tried evaluating the integral of $\displaystyle \int(5-y)-(1/6(y-5)^2)$, with the limits -1 to 5, but was marked incorrect...
That integral $\displaystyle \displaystyle\int_{-1}^5\!\!\bigl(5-y-\tfrac16(y-5)^2\bigr)dy$ looks correct to me. Maybe you evaluated it wrongly. Did you get the answer 6?

4. Originally Posted by Opalg
That integral $\displaystyle \displaystyle\int_{-1}^5\!\!\bigl(5-y-\tfrac16(y-5)^2\bigr)dy$ looks correct to me. Maybe you evaluated it wrongly. Did you get the answer 6?
You're right... I think I made a mistake in the integration of 1/6(y-5)^2.. Thank you

5. Originally Posted by Opalg
That integral $\displaystyle \displaystyle\int_{-1}^5\!\!\bigl(5-y-\tfrac16(y-5)^2\bigr)dy$ looks correct to me. Maybe you evaluated it wrongly. Did you get the answer 6?
Isn't he integrating from y=0 to y=6 ? I graphed the functions and they intersect at (-1,6) and (5,0). You can't plug in x values for y.