## anonymous one year ago The bass of solid S is the region enclosed by the parabola y = 64 - 25x^2 and the x-axis. Cross-sections perpendicular to the y-axis are squares. Find the volume of the described solid S. Where do I even start?!

1. IrishBoy123

I'm on mobile so can't draw but all they're really saying is that the height of the solid z = f(x,y) =64-25x^2 So both width and height of a cross section are driven by the value of x Integrate f(x,y) over the region.

2. IrishBoy123

Actually no need for double integral. Just create a function A(x) for area at any x and integrate that dx.

3. anonymous

So integrate 64-25x^2? @irishboy123

4. IrishBoy123

$$Width = 64-25x^2$$ $$Height = 64-25x^2$$ $$A(x) = W \cdot H = (64-25x^2)^2$$

5. anonymous

Oh really? I got area to be $\frac{ 4(64-x) }{ 25 }$

6. anonymous

And then integrating that from 0 to 1 to get 254/25 as my answer?

7. anonymous

Sorry, x was meant to be y, my bad.

8. IrishBoy123

|dw:1444053041996:dw|

9. anonymous

10. IrishBoy123

drat "perpendicular to" y axis means this..... |dw:1444054108785:dw| $$x = \sqrt{\dfrac{64-y}{25}}$$ $$A(y) = \left[ \sqrt{\dfrac{64-y}{25}} \right]^2$$ $$V = \int\limits_{0}^{64} \; \dfrac{64-y}{25} \; dy = \dfrac{2048}{25}$$

11. anonymous

Oh yeah, that makes sense! Did you accidentally delete your previous reply? :P

12. IrishBoy123

no i thought it had a mistake and i deliberately deleted it and then re-typed it and drew it......again :-) i'm just looking at that web link you produced...........i don't get the 2x part but i'll keep trying.

13. anonymous

Aha, oh okay then :) Me neither to be honest with you, but the next step made sense in my head... Somewhat :p

14. anonymous

Thanks again for your help! Much appreciated :)

15. IrishBoy123

oh i see the 2 is because it is both sides of the y axis....

16. IrishBoy123

which is also the case here...so that factor of 2 goes in as well i did this in my head on mobile and assumed .....wrongly |dw:1444054850625:dw|

17. anonymous

So the answer is actually 8192/25? Go figure :p

18. IrishBoy123

$$A(y) = \left[ \color{red} 2 \sqrt{\dfrac{64-y}{25}} \right]^2 = 4 \sqrt{\dfrac{64-y}{25}}$$ $$V = \color{red} 4\int\limits_{0}^{64} \; \dfrac{64-y}{25} \; dy = 4 \cdot \dfrac{2048}{25}$$

19. IrishBoy123

$=\dfrac{8192}{25}$

20. anonymous

Sweet, mystery solved!

21. IrishBoy123

thank you for your patience, have we licked it?!?!

22. anonymous

I think we have! Always a pleasure working with you Mr. Holmes :p :)

23. IrishBoy123

thank you Dr Watson, and vice versa :-))