## anonymous 5 years ago Sketch the region enclosed by the given curves. Decide whether to integrate with respect to x or y. Draw a typical approximating rectangle and label its height and width. 6x+y^2=13, x=2y find area of the region.

1. amistre64

x = -(1/6)y^2 + 13/6 ; x = 2y

2. amistre64

id integrate with respect to y in the end; maybe :)

3. anonymous

when you integrate with respect to y, do you have the it in terms of x or y ? im a little confused D:

4. amistre64

terms of y

5. amistre64

we need to equate the 2 equations i wrote and find the 'boundarys'

6. amistre64

(-1/6)y^2 -2y +13/6 = 0 will give us the interval between the ys

7. anonymous

8. amistre64

multiply by -6 to normalize this thing right? y^2 +12y -13 = 0

9. anonymous

i think it should be integrated wrt x.check the curve please...which i have given...

10. amistre64

y = -13, and y = 1

11. amistre64

coulda tried to save it as a jpeg ya know :) loads faster

12. amistre64

wrt x is fine; but trickier; wrt y is easier and more straight forward

13. anonymous

for the region left where x=2y intersects the parabola. the y is x/2, and for the right the y=sqrt(13-6x).

14. amistre64

when we integrate this area, we need to subtract one function from the other to get the total area; do we do absolute values? or let the areas lie where they are either pos or neg?

15. anonymous

so the integral would look something like this.. $\int\limits_{-13}^{1} (-1/6)y^2 +13/6-2y)dy$

16. amistre64

i believe so :)

17. anonymous

than you both so much :)

18. anonymous

thank*

19. amistre64

if it turns out negative; just take the absolute value; it simply means you subtracted in the wrong order for example: 5-3 = 2 3-5 = -2; but |-2| = 2

20. anonymous

gotcha. thanks a bunch :)

21. amistre64

youre welcome :)

22. anonymous

it all worked out. :D you rock :)

23. amistre64

:) thnx