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Nesterboy [21]
3 years ago
9

Find the volume of the solid whose base is the region Ix I Iy I < 1 and whose vertical cross sections perpendicular to the y-

axis are semicir- cles (with diameter along the base) g
Mathematics
1 answer:
gavmur [86]3 years ago
4 0

Answer:

volume of the solid = \frac{\pi }{3}

Step-by-step explanation:

As given , the region : |x| + |y| \leq 1

when y \geq 0 ,

|x| + y = 1

⇒|x| = 1 - y

when y ≤ 0 ,

|x| - y = 1

⇒|x| = 1 + y

The given region is the rectangle with sides (0, 5) , (0, -5), (5, 0), (-5, 0)

As given , the vertical cross section are semicircle

As we know that the area of semi circle  = \frac{1}{2}\pi  x^{2}

Now,

Volume = \int\limits^0_-1    {\frac{\pi }{2} (1+y)^{2}  } \, dy + \int\limits^1_0    {\frac{\pi }{2} (1-y)^{2}  } \, dy

            = \frac{\pi }{2} (\frac{(1+y)^{3} }{3} ) - \frac{\pi }{2} (\frac{(1-y)^{3} }{3} )

           = \frac{\pi }{6}[1-0] - \frac{\pi }{6}[0-1]

           = \frac{\pi }{6} + \frac{\pi }{6} = 2\frac{\pi }{6} = \frac{\pi }{3}

⇒volume of the solid = \frac{\pi }{3}

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