Question

A wedge is cut out of a circular cylinder of radius 4 by two planes. One is perpendicular to the axis of the cylinder; the other intersects the first at an angle of 30 along a diameter of the cylinder. Find the volume of the wedge.

Answer 1

Answer:

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Answer 2

Volume of Wedge  = $\frac{128}{3\sqrt{3} }$

Step-by-step explanation:

Here, Along  x -axis the diameter where the point meets , the the base of solid is a semicircle  with equation y  =$\sqrt{16 - x^{2} }$ , -4 ≤ x ≤ 4

A cross section perpendicular to the x-axis at a distance x from the origin in a ΔABC  

Base  y = $\sqrt{16 -x^{2} }$

height = y tan 30°....(i)

Put y value in this (i)

height  = $\sqrt{16-x^{2} }$ /$\sqrt{3}$

Area of cross-sectional A(x) = 1/2 . 1/$\sqrt{3}$  $\sqrt{16-x^{2} }$ $\sqrt{16 -x^{2} }$

                                       A(x) = 1/2√3 (16 - x²)

Volume of wedge = $\int_{-4}^{4}A(x)dx$ $= \int_{-4}^{4} \frac{16-x^{2} }{2\sqrt{3} }$ dx

= $\frac{1}{\sqrt{3} }$ $\int_{0}^{4}(16-x^{2})dx$

= $\frac{1}{\sqrt{3} }$  [ 16 x - $\frac{x^{3} }{3}$ ]₀⁴

= $\frac{128 }{3\sqrt{3} }$