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A frustum of a cone is the portion of the cone bounded between the circular base and a plane parallel to the base. With dimensions are indicated, show that the volume of the frustum of the cone is V=13R2H13rh2
SHOW FULL SOLUTION AND EXPLAIN. INTEGRAL CALCULUS. A. Using the disk method, evaluate the volume of the solid generated when the area bounded by the parabola y^2=-4x+4 and the y axis is revolved about the y axis. B. Set up the volume integral in (A) if the area is rotated about x=1 using a horizontal element.
Solve by integration: A frustum of cone is made by rotating about Ox the area bounded by the lines y=x, x=0, x=a and x=b, with b>a. What is its volume?
The sphere x 2 + y 2 + z 2= 2a 2 and the region inside the cylinder x 2 + y 2 = a 2 Find the volume.
Use Pappus's Theorem to find the volume of the solid obtained when the region bounded by y=4x2, y=0, and x=1, whose centroid is 34,310, is revolved about the y-axis. Check your answer by using the method of cylindrical shells. What is the volume of the solid in cubic units?
Use cylindrical coordinates to find the volume of the region bounded by the plane z = 0 and the hyperboloid z= √37-√ 1 + x² + y² Set up the triple integral using cylindrical coordinates that should be used to find the volume of the region as as possible. Use increasing limits of integration. 10*** dz dr de RAMON ⁰77 0
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