The base of a solid is a circle with radius 6. Each cross section perpendicular to a fixed diameter of the base is semicircular.
If the circle is centered at the origin and the fixed diameter lies on the x-axis, find the cross-section area A(x).
A(x) =
Find the volume V of the solid.
V =
A(x) = π(6^2 - x^2)
V = (1/2)π∫x=-6x=6(6^2 - x^2)dx
V = (1/2)π[6x(6^2 - x^2) - (1/3)(6^2 - x^2)^3]|x=-6 to x=6
V = (1/2)π[6(36 - 36) - (1/3)(36 - 36)^3]
V = 0
V=0?
Surely that cannot be true!
Of course that garbage you wrote prolly didn't help ...
A(x) = π/2 (36-x^2)
using symmetry,
V = 2∫[0,6] π/2 (36-x^2) dx
= ∫[0,6] π (36-x^2) dx
= 144π
To find the cross-section area A(x), we need to find the radius of each semicircle at a given x-coordinate.
Since the circle is centered at the origin and the fixed diameter lies on the x-axis, the x-coordinate of any point on the circle is equal to its radius.
We know that the radius of the base circle is 6. Therefore, the radius of each semicircle at a given x-coordinate is equal to the x-coordinate itself.
Hence, A(x) = the area of a semicircle with radius x.
The formula for the area of a semicircle is A = (π * r^2) / 2.
Plugging in our radius x, we get A(x) = (π * x^2) / 2.
Now, let's find the volume V using integral calculus.
The volume of the solid can be determined by integrating the cross-section area A(x) with respect to x from the leftmost point to the rightmost point of the solid.
The leftmost point of the solid is the negative of the radius, which is -6, and the rightmost point is the radius itself, which is 6.
Therefore, V = ∫[from -6 to 6] A(x) dx.
Substituting the formula for the cross-section area A(x), we get V = ∫[from -6 to 6] (π * x^2) / 2 dx.
Evaluating this integral will give us the volume V of the solid.
To find the cross-section area A(x), we need to consider that each cross-section perpendicular to a fixed diameter of the base is semicircular.
Since the circle is centered at the origin and the fixed diameter lies on the x-axis, we can represent the equation of the circle as x^2 + y^2 = 6^2.
The equation of a semicircle centered at the origin with radius r is given by y = √(r^2 - x^2).
In this case, the radius of the semicircle is 6, so the equation becomes y = √(6^2 - x^2).
To find the cross-section area A(x), we need to integrate this equation to get the area of the semicircle.
A(x) = ∫[−6, 6] √(6^2 - x^2) dx
To find the volume V of the solid, we need to integrate the cross-section area A(x) over the entire length of the solid, which is the interval [-6, 6].
V = ∫[−6, 6] A(x) dx
To evaluate these integrals, we can use techniques such as substitution or trigonometric substitution, depending on the complexity of the calculations.