Question

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

Answer 1

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 = $\frac{\pi }{3}$

Answer 2

Volume of the solid is the value of the capacity of the object that can be hold by it. Thus the value of the volume of the given solid is,

$V=\dfrac{\pi}{3}$

Given information-

The vertical cross sections of the solid is perpendicular to the y-axis.

The region of the base is,

$|x|+ |y|\leq 1$

Volume

Volume of the solid is the value of the capacity of the object that can be hold by it.

When the value of the variable y is greater than or equal to zero than the value of the x is,

$|x|=1-y$

When the value of the variable y is less than or equal to zero than the value of the x is,

$|x|=1+y$.

Given inequality makes a square with the corners (1,0), (0,1),(-1,0),(0,-1).

Now the limit for the given solid is between 0 to 1 and area is equal to the semicircle which is,

$A=\dfrac{\pi x^2}{2}$

Thus the volume of the solid whose area bounded by the region 1 to -1 and vertical cross section is semicircle can be given as,

$V=\int\limits^1_0 {\dfrac{\pi x^2}{2} } \, dx$

$V=\dfrac{\pi}{2} [\;\;\int\limits^1_0 {(1+y)^2} \, dy +\int\limits^1_0 {(1-y)^2} \, dy ]$

$V=\dfrac{\pi}{2} [\;\;\dfrac{\int\limits^1_0 {(1+y)^3}}{3} \, +\dfrac{\int\limits^1_0 {(1-y)^3} \, }{3} ]$

$V=\dfrac{\pi}{6} [\;\;\ {(1-0)^3} \, +{(0-1)^3} \, ]$

$V=\dfrac{\pi}{3}$

Thus the value of the volume of the given solid is,

$V=\dfrac{\pi}{3}$

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