Asked by Louis
The base of a solid in the xy-plane is the circle x^2+y^2 = 16. Cross sections of the solid perpendicular to the y-axis are semicircles. What is the volume, in cubic units, of the solid?
a. 128π/3
b. 512π/3
c. 32π/3
d. 2π/3
a. 128π/3
b. 512π/3
c. 32π/3
d. 2π/3
Answers
Answered by
bobpursley
<<Cross sections of the solid perpendicular to the y-axis are semicircles>>
Somehow, I am not visulizing this. so the base of the solid is in the xy plane, so its altitude must be in the k direction. For the k direction to be semicircles, the shape of the solid has to be a semisphere.
the radius is 4. The volume of the semisphere then is 1/2 the sphere of radius 4,
volume= 1/2 * 4/3 PI 4^3
Somehow, I am not visulizing this. so the base of the solid is in the xy plane, so its altitude must be in the k direction. For the k direction to be semicircles, the shape of the solid has to be a semisphere.
the radius is 4. The volume of the semisphere then is 1/2 the sphere of radius 4,
volume= 1/2 * 4/3 PI 4^3
Answered by
Steve
Each semicircle resting on the x-y plane has radius x=√(16-y^2). Adding up all the slices, and using symmetry, we have
v = 2∫[0,4] 1/2 πr^2 dy
= π∫[0,4] (16-y^2) dy = 128π/3
bobpursley's solution is much more intuitive and geometric, though, eh?
v = 2∫[0,4] 1/2 πr^2 dy
= π∫[0,4] (16-y^2) dy = 128π/3
bobpursley's solution is much more intuitive and geometric, though, eh?
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