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:

volume of the solid = [tex]\frac{\pi }{3}[/tex]

Step-by-step explanation:

As given , the region : |x| + |y| [tex]\leq[/tex] 1

when y [tex]\geq[/tex] 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  = [tex]\frac{1}{2}[/tex][tex]\pi x^{2}[/tex]

Now,

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

            = [tex]\frac{\pi }{2} (\frac{(1+y)^{3} }{3} )[/tex] - [tex]\frac{\pi }{2} (\frac{(1-y)^{3} }{3} )[/tex]

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

           = [tex]\frac{\pi }{6} + \frac{\pi }{6} = 2\frac{\pi }{6} = \frac{\pi }{3}[/tex]

⇒volume of the solid = [tex]\frac{\pi }{3}[/tex]

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,

[tex]V=\dfrac{\pi}{3} [/tex]

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

The region of the base is,

[tex]|x|+ |y|\leq 1[/tex]

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,

[tex]|x|=1-y[/tex]

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

[tex]|x|=1+y[/tex].

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,

[tex]A=\dfrac{\pi x^2}{2} [/tex]

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,

[tex]V=\int\limits^1_0 {\dfrac{\pi x^2}{2} } \, dx [/tex]

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

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

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

[tex]V=\dfrac{\pi}{3} [/tex]

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

[tex]V=\dfrac{\pi}{3} [/tex]

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