Answer :
The required volume of the solid whose base is a circular disk is [tex]\frac{1024}{3}r^{3}[/tex].
Explain the volume of solid.
The volume of a strong is the proportion of how much space an item takes up. It is estimated by the quantity of unit shapes it takes to top off the strong. Including the unit shapes in the strong, we have 30 unit blocks, so the volume is: 2 units 3 units 5 units = 30 cubic units.
According to question:
We have circular place, with origin at center
x^2 + ^2 = 16r^2
Taking into account that piece of the strong in the first octane, with the square cross-segments running lined up with the x-z pivot, the volume of a natural cross area is:
dv = x(2x)dy = 2(16r^2 - y^2)dy
On integrating it
V = 2∫₀^(4r) { (16r^2 - y^2)dy}
[tex]2[16r^{2}y-\frac{y^{3} }{3}]_{0}[/tex] = [tex]\frac{256}{3}r^{3}[/tex]
This volume is one quadrant,
So, total volume is = [tex]\frac{1024}{3}r^{3}[/tex]
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