Respuesta :
Answer: V = [tex]\frac{64}{3}\pi[/tex]
Step-by-step explanation: A solid formed by revolving the region about the x-axis can be considered to have a thin vertical strip with thickness Δx and height y = f(x). The strip creates a circular disk with volume:
V = [tex]\pi. y^{2}.[/tex]Δx
Using the Disc Method, it is possible to calculate all the volume of these strips, giving the volume of the revolved solid:
V = [tex]\int\limits^a_b {\pi. y^{2} } \, dx[/tex]
Then, for the region generated by y = - x + 4:
V = [tex]\int\limits^4_0 {\pi.(-x+4)^{2} } \, dx[/tex]
V = [tex]\pi.\int\limits^4_0 {(x^{2}-8x+16)} \, dx[/tex]
V = [tex]\pi.(\frac{x^{3}}{3}-4x^{2}+16x )[/tex]
V = [tex]\pi.(\frac{4^{3}}{3}-4.4^{2}+16.4 - 0 )[/tex]
V = [tex]\frac{64}{3}.\pi[/tex]
The volume of the revolved region is V = [tex]\frac{64}{3}.\pi[/tex]
The evaluation of the definite integral that represents the volume of the solid is [tex]\mathbf{\dfrac{64 \pi}{3}}[/tex]
Using the Disk Method to determine the volume of a solid formed by revolving the region about the x-axis and the interval [a, b]; we have:
[tex]\mathbf{V = \int ^b_a \pi (y)^2 \ dx}[/tex]
where;
- b = 4
- a = 0
[tex]\mathbf{V = \int ^4_0 \pi \Big[-x +4 \Big]^2 \ dx}[/tex]
[tex]\mathbf{V =\pi \int ^4_0\Big[-x^2 -8x+16 \Big] \ dx}[/tex]
[tex]\mathbf{V =\pi \Big[\dfrac{x^3}{3} -4x^2+16x \Big]^4_0 \ dx}[/tex]
[tex]\mathbf{V =\pi \Big[\dfrac{4^3}{3} -4(4)^2+16(4) -0 \Big]}[/tex]
[tex]\mathbf{V =\dfrac{64 \pi}{3}}[/tex]
Therefore, we can conclude that the evaluation of the definite integral that represents the volume of the solid is [tex]\mathbf{\dfrac{64 \pi}{3}}[/tex]
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