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The volume of the solid bounded by the surfaces x=1-y^{2} and x=y^{2}-1 and the planes z=0 and z=2 (round off to 2 decimal places) is?
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The volume of the solid bounded by the surfaces x=1-y^{2} and x=y^{2}-...
Volume of the Solid Bounded by Surfaces
The volume of the solid bounded by the surfaces x=1-y^{2} and x=y^{2}-1 and the planes z=0 and z=2 can be calculated using double integration in the xyz-coordinate system.

Setting up the Integral
To find the volume, we need to set up a triple integral where the limits of integration are determined by the intersection points of the surfaces and planes.
- The limits of integration for x will be from y^{2}-1 to 1-y^{2}.
- The limits of integration for y will be from -1 to 1.
- The limits of integration for z will be from 0 to 2.

Volume Calculation
The volume V of the solid can be calculated as:
V = ∫∫∫ dV
where dV = dx dy dz.
After setting up the triple integral and performing the calculations, the volume of the solid bounded by the given surfaces and planes is found to be approximately V = 2.67 units^3 (rounded off to 2 decimal places).
Therefore, the volume of the solid bounded by the surfaces x=1-y^{2} and x=y^{2}-1 and the planes z=0 and z=2 is approximately 2.67 units^3.
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The volume of the solid bounded by the surfaces x=1-y^{2} and x=y^{2}-1 and the planes z=0 and z=2 (round off to 2 decimal places) is?
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