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

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IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1I'm on mobile so can't draw but all they're really saying is that the height of the solid z = f(x,y) =6425x^2 So both width and height of a cross section are driven by the value of x Integrate f(x,y) over the region.

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1Actually no need for double integral. Just create a function A(x) for area at any x and integrate that dx.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So integrate 6425x^2? @irishboy123

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1\(Width = 6425x^2\) \(Height = 6425x^2\) \(A(x) = W \cdot H = (6425x^2)^2\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Oh really? I got area to be \[\frac{ 4(64x) }{ 25 }\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0And then integrating that from 0 to 1 to get 254/25 as my answer?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Sorry, x was meant to be y, my bad.

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1dw:1444053041996:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I followed my friend's suggestion and had a look at this link: http://slader.com/textbook/9780538498845stewartcalculusinternationaledition7thedition/362/exercises/58/

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1drat "perpendicular to" y axis means this..... dw:1444054108785:dw \(x = \sqrt{\dfrac{64y}{25}}\) \(A(y) = \left[ \sqrt{\dfrac{64y}{25}} \right]^2 \) \(V = \int\limits_{0}^{64} \; \dfrac{64y}{25} \; dy = \dfrac{2048}{25}\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Oh yeah, that makes sense! Did you accidentally delete your previous reply? :P

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1no i thought it had a mistake and i deliberately deleted it and then retyped 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.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Aha, oh okay then :) Me neither to be honest with you, but the next step made sense in my head... Somewhat :p

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Thanks again for your help! Much appreciated :)

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1oh i see the 2 is because it is both sides of the y axis....

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1which 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

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So the answer is actually 8192/25? Go figure :p

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1\(A(y) = \left[ \color{red} 2 \sqrt{\dfrac{64y}{25}} \right]^2 = 4 \sqrt{\dfrac{64y}{25}} \) \(V = \color{red} 4\int\limits_{0}^{64} \; \dfrac{64y}{25} \; dy = 4 \cdot \dfrac{2048}{25}\)

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1\[=\dfrac{8192}{25}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Sweet, mystery solved!

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1thank you for your patience, have we licked it?!?!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I think we have! Always a pleasure working with you Mr. Holmes :p :)

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1thank you Dr Watson, and vice versa :))
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