## anonymous 3 years ago find the volume of the given solid: under the surface z=2x+y^2 and above the triangle vertices (1,1),(4,1) and (1,2).

1. anonymous

|dw:1360708771803:dw|

2. TuringTest

what is the equation of this part of the area?|dw:1360709404014:dw|

3. anonymous

y=2x?

4. TuringTest

no,you can see that the slope from x=1 to x=4 is down for that part of the graph, so the coefficient of x will be negative you should use point-slope form way back from algebra I to do this|dw:1360709566120:dw|

5. anonymous

@TuringTest yes i can see that

6. anonymous

so -1/3?

7. TuringTest

yes, so now use point-slope form on either point $$(x_0,y_0)$$ to get the equation of the line $y-y_0=-\frac13(x-x_0)$

8. anonymous

y-1 = -1/3 (x-1)?

9. TuringTest

no, because (1,1) is not one of the points on the diagonal line

10. anonymous

ok

11. anonymous

the only one that i can use the (1,2) and (1,4)

12. TuringTest

correct, we are using point-slope form on the diagonal line, so we can only get the equation from points that lie $$on$$ the line.

13. anonymous

y-2= -1/3 (x-1)

14. TuringTest

yes, now solve for y...

15. anonymous

y= -1/3x+1/3 (+2) y=-1/3 x+7/3

16. TuringTest

yes, and now you need the line that bounds the region below:|dw:1360710827125:dw|

17. anonymous

same using slope

18. TuringTest

er, yeah you could, but you should really be able to eyeball this one. Note here that all we have been doing so far is basic algebra to find the equations of the bounds of the region. Just because you are in calc2 does not mean you can forget the basics!! On the contrary, here is where you actually need them most.

19. anonymous

really is calc 3

20. TuringTest

Well, what some people call cal3 others call calc2. MIT for instance only has 2 basic calc classes: single and multivariable. But enough semantics, what's the equation of the bottom line?

21. anonymous

@TuringTest is 1?

22. TuringTest

y=1, yes so y is bound by the two equations we found, so they will be the bounds for the inner integral, which will be with respect to x. what are the bounds on x, which we will use for the outer integral.

23. anonymous

x=1, 4 and y= 1 , 1/3x+7/3?

24. TuringTest

yes :)

25. anonymous

$\int\limits_{1}^{4} \int\limits_{1}^{1/3x+7/3} xy$ dy dx

26. TuringTest

the integrand is the function for z, so use that, not xy gotta go, happy integrating!

27. anonymous

@TuringTest thank you so much

28. anonymous

sorry is not z=2x+y^2 i just mistype the book is really saying xy