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?!

Question

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?!

irishboy123 · Answer

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.

irishboy123 · Answer

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

clarence · Answer

So integrate 64-25x^2? @irishboy123

irishboy123 · Answer

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

clarence · Answer

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

clarence · Answer

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

clarence · Answer

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

irishboy123 · Answer

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clarence · Answer

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/

irishboy123 · Answer

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  $

$V = \int\limits_{0}^{64} \; \dfrac{64-y}{25} \; dy = \dfrac{2048}{25}$