Volume of a solid bound by region work shown

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SUMMARY

The volume of a solid bounded by the region defined by the curve y = 2*sqrt(sin(x)) and the x-axis, where x is in the interval [0, π/2], is determined using square cross-sections perpendicular to the x-axis. The area function A(x) is calculated as A(x) = [2*sqrt(sin(x))]^2, leading to the integral of A(x) from 0 to π/2 to find the volume. The correct integral setup is crucial, as the area must account for the factor of 2 from the original equation. The final volume can be computed using the definite integral of the area function.

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johnq2k7
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The base of a solid is the region bounded by y= 2*sqrt(sin(x)) and the x-axis, with x an element of [0, (pi/2)]. Find the volume of the solid, given that the cross sections perpendicular to the x-axis are squares.

Work Shown:

cross sections are squares:

therefore A(x) is not equal to Pi*r^2 rather l^2

therefore A(x)= [sqrt(sin(x)]^2

therefore integral of A(x)dx of x-values from 0 to Pi/2 should provide the answer

I think my approach is wrong and my integral for A(x) dx is wrong as well please help!
 
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Hi johnq2k7! :smile:

(have a pi: π :wink:)

Seems ok (except you seem to have droppoed the original 2) …

the general rule is to specify the slices you're dividing it into … in this case, slices of thickness dx and area y2. :wink:
 

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