Answer:
Step-by-step explanation:
Let's give this a go here. The volume formula for the shell method while rotating about a horizontal line is
[tex]V=2\pi \int\limits^a_b {p(y)h(y)} \, dy[/tex]
where p(y) is the distance from the axis of rotation (the x-axis) to the center of the solid. This is a positive distance and it is just y.
h(y) is the horizontal height of the function. Our function starts at x = 0 and ends at the function itself, so h(y) = 3 + y^2.
In the shell method when rotating about a horizontal line, we need to use x = y equations, and y-intervals. Setting up our integral then:
[tex]V=2\pi \int\limits^3_2 {y(3+y^2)} \, dy[/tex]
We can simplify this a bit by distributing the y into the parenthesis:
[tex]V=2\pi \int\limits^3_2 {3y+y^3} \, dy[/tex]
Integrating gives us
[tex]V=2\pi|\frac{3y^2}{2}+\frac{y^4}{4}|[/tex] from 2 to 3
Using the First Fundamental Theorem of Calculus:
[tex]V=2\pi[\frac{135}{4}-\frac{40}{4}][/tex] which simplifies down to
[tex]V=\frac{95\pi}{2}[/tex]
We have that the volume V of the solid obtained by rotating the region bounded by the given curves about the x-axis is
[tex]V=149.2[/tex]
From the question we are told
x = 3 + y2,
x = 0,
y = 2,
y = 3
Given a sketch of the Graph of x against y
Generally the equation for the volume is mathematically given as
[tex]V=\int_2^3(2\pi y)*(3+y^2)dy \\\\V=2 \pi \int^3_2(3y+y^3)dy\\\\V=2\pi {\frac{3y^2}{2}+\frac{y^4}{4}}_2^3\\\\V=2 \pi(\frac{95}{4})[/tex]
[tex]V=149.2[/tex]
Therefore
the volume V of the solid obtained by rotating the region bounded by the given curves about the x-axis is
[tex]V=149.2[/tex]
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