## richyw Group Title Volume by DOUBLE integral. Still struggling with the same double integral problem! Test is in two hours! Can someone check my attempt? one year ago one year ago

1. richyw Group Title

set up the double integral to evaluate the volume of the solid bounded by $$y=x^2$$, $$z=3y$$ and $$z=2+y$$

2. richyw Group Title

I have set it up like this. does it look correct? $V=\int^1_0\int^{2+y}_{3y}x^2\,dzdx$

3. amistre64 Group Title

how did you determine the solid that is bounded?

4. richyw Group Title

I plugged $$y=x^2$$ into $$z=3y$$ and $$z=2+y$$. And found $$x=\pm 1$$.

5. richyw Group Title

I am very confused, all of the text book questions I breezed through and this one has me second guessing...

6. amistre64 Group Title

|dw:1352407804481:dw|

7. amistre64 Group Title

im thinking the solid is something like this; a y=x^2 cylindar that hits the z=3y plane and the z=y+2 plane; giving us 2 seperate volumes to calculate

8. amistre64 Group Title

|dw:1352408012819:dw|

9. amistre64 Group Title

its been a while so i gotta wonder if im interpreting the shape correctly

10. amistre64 Group Title

|dw:1352408252457:dw| or does it look more like a wedge of an orange

11. amistre64 Group Title

the wedge looks more "bounded" by the 3 surfaces to me

12. amistre64 Group Title

so, if we use half of this, it being symmetric in shape; x = 0 to 1 y = x^2 to 1 z = 3y to 2+y

13. amistre64 Group Title

$\huge \int_{x=0}^{x=1}~\int_{y=x^2}^{y=1}~\int_{z=3y}^{z=2+y}~~dz~dy~dx$

14. richyw Group Title

yeah it has to be a double integral though. that's my huge problem. I'm missing something...

15. amistre64 Group Title

solve the inner integral and it turns into a double :/

16. richyw Group Title

I have posted it here before. furthermore I haven't gotten a DI to equal the result of that TI.

17. richyw Group Title

yup that's what I might do on the exam haha.

18. amistre64 Group Title

if you wanna move about the limits of integrations to form different forms of the same TI into to create different forms of the DI ... that would prolly help work out some issues i think the problem is that is is not bounded by a usual plane

19. amistre64 Group Title

do you have an answer key to check with?

20. richyw Group Title

no

21. richyw Group Title

the cylinder has vertical sides, so I think this is correct.

22. richyw Group Title

anyways thanks a lot!

23. amistre64 Group Title

$\int_{0}^{1}~\int_{x^2}^{1}~\int_{3y}^{2+y}~~dz~dy~dx$ $\int_{0}^{1}~\int_{x^2}^{1}(2+y-3y)~dy~dx$ $\int_{0}^{1}~\int_{x^2}^{1}2-2y~dy~dx$ $2\int_{0}^{1}~\int_{x^2}^{1}1-y~dy~dx$ $2\int_{0}^{1}~(1-\frac12)-(x^2-\frac12(x^2)^2)~dx$ $2\int_{0}^{1}~\frac12-x^2+\frac12x^4~dx$

24. amistre64 Group Title

and then double that result since thats only half of it :) otherwise; use each z as a cap over the y=x^2 and subtract the lower cap volume from the upper cap volume

25. amistre64 Group Title

$2\left(2\int_{0}^{1}~\frac12-x^2+\frac12x^4~dx\right)$ $2\left(\int_{0}^{1}~1-2x^2+x^4~dx\right)$ $\int_{0}^{1}~2-4x^2+2x^4~dx$ $2-\frac43+\frac25=\frac{16}{15}$ maybe :)

26. richyw Group Title

thanks a lot. sorry if I wasn't too active. trying to cram! I'll take a good look at it tonight though and figure it out for sure.

27. amistre64 Group Title

good luck :)