The volume of the solid bounded above by the parabolic cylinder is 191π/64.
Given:
the Parabolic cylinder z=1−y^2 and below the plane 3x+3y+z+10=0 and on the sides of circular cylinder x^2+y^2−x=0.
[tex]x^2+y^2-x=(x-\frac12)^2 +y^2-\frac14=0[/tex]
Recenter the circle with x−1/2=u and integrate the volume over the disk
u^2+y^2 = 1/4.
[tex]\int_{u^2+y^2 < \frac14} [(1-y^2)-( -2x-3y-10)][/tex]
[tex]=\int_{u^2+y^2 < \frac14} (2u+3y-y^2 +12)[/tex]
[tex]=\int_{u^2+y^2 < \frac14} (12-y^2)[/tex]
[tex]=\int_0^{2\pi}\int_0^{1/2} (12-r^2 \sin^2\theta)rdr d\theta =\frac{191\pi}{64}\\[/tex]
Therefore The volume of the solid bounded above by the parabolic cylinder is 191π/64.
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