Double Integrals: Volume of a Solid in the First Octant

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SUMMARY

The volume of the solid in the first octant bounded by the cylinder defined by the equation y² + z² = 9 and the plane x = 6 can be calculated without integration. The area of one-quarter of the circle formed by the cylinder is π(3²)/4, which equals 9π/4. Multiplying this area by the length of the solid along the x-axis (6) yields a total volume of 54π/4 or 13.5π. For those who prefer integration, the double integral can be set up as ∫ from y=0 to 3 of ∫ from z=0 to √(9 - y²) dz dy.

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How would you find the volume in the first octant of the solid bounded by the cylinder y2+z2=9 and x=6?

I tried doing a double integral, both from 0 to 3 of (y2+z2)dydz then multiply by 6, but I keep getting 0.
 
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y and z both going from 0 to 3 gives a rectangle.

[itex]y^2+ z^2= 9[/itex] is a circle of radius 3. What is the area of that circle? What is the area of 1/4 of that circle? Since x= 6 is parallel to the yz-plane, just multiply the area of 1/4 the circle by 6. No integration required.

(If you really want to do an integration, then you could take y from 0 to 3 but then z, for each y, must be from 0 to [itex]z= \sqrt{9- y^2}[/itex]. That is, the area is
[tex]\int_{y=0}^3\int_{z= 0}^{\sqrt{9- y^2}}dzdy\right)[/tex]
 
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