Respuesta :
find where they intersect
f(x)=g(x) at
ok
9x=x^2+6x
0=x^2-3x
intersect at x=0 and x=3
which one is on 'top'
f(2)=18
g(2)=16
f(x) is on top
so it is
(we don't need to find r)
remember, the volume rotated around x axis when f(x) is top curve and g(x) is botttom and bounds are a and b is
[tex] \int\limits^a_b {(f(x))^2-(g(x))^2} \, dx [/tex]
so
f(x)=9x
g(x)=x^2+6x
the bounds are 0 and 3
[tex] \int\limits^3_0 {(9x)^2-(x^2+6x)^2} \, dx [/tex]=v
[tex] \int\limits^3_0 {x^4+12x^3+117x^2} \, dx [/tex]=v
[tex][ \frac{x^5}{5}+3x^4+ 39x^3 ]^3_0[/tex]=v
[tex] \frac{243}{5}+1296=v [/tex]
the volume is 1344.6 cubic units
f(x)=g(x) at
ok
9x=x^2+6x
0=x^2-3x
intersect at x=0 and x=3
which one is on 'top'
f(2)=18
g(2)=16
f(x) is on top
so it is
(we don't need to find r)
remember, the volume rotated around x axis when f(x) is top curve and g(x) is botttom and bounds are a and b is
[tex] \int\limits^a_b {(f(x))^2-(g(x))^2} \, dx [/tex]
so
f(x)=9x
g(x)=x^2+6x
the bounds are 0 and 3
[tex] \int\limits^3_0 {(9x)^2-(x^2+6x)^2} \, dx [/tex]=v
[tex] \int\limits^3_0 {x^4+12x^3+117x^2} \, dx [/tex]=v
[tex][ \frac{x^5}{5}+3x^4+ 39x^3 ]^3_0[/tex]=v
[tex] \frac{243}{5}+1296=v [/tex]
the volume is 1344.6 cubic units
The volume of the solid generated by revolving about the x-axis is [tex]\frac{567}{5}\pi[/tex] cubic units.
How to generate to a solid of revolution of a region around the x-axis
The points of intersection between [tex]f(x)[/tex] and [tex]g(x)[/tex] are [tex](0, 0)[/tex] and [tex](3, 27)[/tex]. The formula of a solid of revolution generated around the x-axis is represented by the following integral formula:
[tex]V = \pi \int\limits^{b}_{a} {\{[f(x)]^{2}-[g(x)]^{2}\}} \, dx[/tex] (1)
If we know that [tex]a = 0[/tex], [tex]b = 3[/tex], [tex]f(x) = 9\cdot x[/tex] and [tex]g(x) = x^{2}+6\cdot x[/tex], then the volumen of the solid about the x-axis is:
[tex]V = \pi \int\limits^{3}_{0} {[9\cdot x]^{2}} \, dx -\pi \int\limits^{3}_{0} {[x^{2}+6\cdot x]^{2}} \, dx[/tex]
[tex]V = 81\pi \int\limits^3_0 {x^{2}} \, dx - \pi \int\limits^{3}_{0} {[x^{4}+12\cdot x^{3}+36\cdot x^{2}]} \, dx[/tex]
[tex]V = -\pi \int\limits^{3}_{0} {x^{4}} \, dx - 12\pi \int\limits^{3}_{0} {x^{3}} \, dx +45\pi\int\limits^{3}_{0} {x^{2}} \, dx[/tex]
[tex]V = -\frac{\pi}{5}\cdot (3^{5}-0^{5}) - \frac{12\pi}{4}\cdot (3^{4}-0^{4}) +\frac{45\pi}{3}\cdot (3^{3}-0^{3})[/tex]
[tex]V =\frac{567}{5}\pi[/tex]
The volume of the solid generated by revolving about the x-axis is [tex]\frac{567}{5}\pi[/tex] cubic units. [tex]\blacksquare[/tex]
To learn more on solids of revolution, we kindly invite to check this verified question: https://brainly.com/question/338504