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?!

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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?!

Mathematics
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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.
Actually no need for double integral. Just create a function A(x) for area at any x and integrate that dx.
So integrate 64-25x^2? @irishboy123

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\(Width = 64-25x^2\) \(Height = 64-25x^2\) \(A(x) = W \cdot H = (64-25x^2)^2\)
Oh really? I got area to be \[\frac{ 4(64-x) }{ 25 }\]
And then integrating that from 0 to 1 to get 254/25 as my answer?
Sorry, x was meant to be y, my bad.
|dw:1444053041996:dw|
I followed my friend's suggestion and had a look at this link: http://slader.com/textbook/9780538498845-stewart-calculus-international-edition-7th-edition/362/exercises/58/
drat "perpendicular to" y axis means this..... |dw:1444054108785:dw| \(x = \sqrt{\dfrac{64-y}{25}}\) \(A(y) = \left[ \sqrt{\dfrac{64-y}{25}} \right]^2 \) \(V = \int\limits_{0}^{64} \; \dfrac{64-y}{25} \; dy = \dfrac{2048}{25}\)
Oh yeah, that makes sense! Did you accidentally delete your previous reply? :P
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.
Aha, oh okay then :) Me neither to be honest with you, but the next step made sense in my head... Somewhat :p
Thanks again for your help! Much appreciated :)
oh i see the 2 is because it is both sides of the y axis....
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|
So the answer is actually 8192/25? Go figure :p
\(A(y) = \left[ \color{red} 2 \sqrt{\dfrac{64-y}{25}} \right]^2 = 4 \sqrt{\dfrac{64-y}{25}} \) \(V = \color{red} 4\int\limits_{0}^{64} \; \dfrac{64-y}{25} \; dy = 4 \cdot \dfrac{2048}{25}\)
\[=\dfrac{8192}{25}\]
Sweet, mystery solved!
thank you for your patience, have we licked it?!?!
I think we have! Always a pleasure working with you Mr. Holmes :p :)
thank you Dr Watson, and vice versa :-))

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