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Find the volume v of the solid obtained by rotating the region bounded by the given curves about the specified line. y = x + 1, y = 0, x = 0, x = 2; about the x-axis

Sagot :

The volume of a solid is  [tex]\frac{26}{3} \pi[/tex].

Given

The given curves about the specified line. y = x + 1, y = 0, x = 0, x = 2; about the x-axis

Curve is y = x + 1

Line is y = 0

We have to find out the volume v of the solid obtained by rotating the region bounded by these curves.

If the region bounded above by the graph of f, below by the x-axis, and on the sides by x=a and x=b is revolved about the x-axis, the volume V of the generated solid is given by [tex]V = \pi \int\limits^b_a {(f(x))^{2} } \, dx[/tex]. We can also obtain solids by revolving curves about the y-axis.

Volume of a solid:

According to washer method:

[tex]V = \pi \int\limits^b_a {(f(x))^{2} } \, dx[/tex]

Using washer method, where a=0 and b=2, we get

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

= [tex]\pi[ \frac{(x+1)^{3} }{3} ]0 \ to \ 2[/tex]

= [tex]\pi [\frac{(3+1)^{3} }{3} -\frac{(0+1)^{3} }{3}][/tex]

= [tex]\pi [\frac{27}{3} -\frac{1}{3} ][/tex]

= [tex]\pi [\frac{26}{3}][/tex]

= [tex]\frac{26}{3} \pi[/tex]

Therefore the volume of a solid is  [tex]\frac{26}{3} \pi[/tex].

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