The base of a solid in the xy-plane is the circle x^2+y^2 = 16. Cross sections of the solid perpendicular to the y-axis are semicircles. What is the volume, in cubic units, of the solid?
a. 128π/3
b. 512π/3
c. 32π/3
d. 2π/3
Each semicircle resting on the x-y plane has radius x=√(16-y^2). Adding up all the slices, and using symmetry, we have
v = 2∫[0,4] 1/2 πr^2 dy
= π∫[0,4] (16-y^2) dy = 128π/3
bobpursley's solution is much more intuitive and geometric, though, eh?
<<Cross sections of the solid perpendicular to the y-axis are semicircles>>
Somehow, I am not visulizing this. so the base of the solid is in the xy plane, so its altitude must be in the k direction. For the k direction to be semicircles, the shape of the solid has to be a semisphere.
the radius is 4. The volume of the semisphere then is 1/2 the sphere of radius 4,
volume= 1/2 * 4/3 PI 4^3
To find the volume of the solid, we can integrate the area of the cross sections as we move along the y-axis.
The given solid has cross sections that are semicircles perpendicular to the y-axis. From the equation of the base circle, we can see that the radius of each semicircle is 4 (since the equation of a circle centered at the origin is (x-h)^2 + (y-k)^2 = r^2, where (h,k) is the center of the circle).
Let's integrate the area of each semicircle as we move from y = -4 to y = 4 (since the radius of the base circle is 4):
∫[from -4 to 4] π(1/2 * (4)^2) dy
Simplifying this integral, we get:
∫[from -4 to 4] 8π dy
Integrating, we have:
[8πy] [from -4 to 4]
Substituting the limits:
8π(4) - 8π(-4)
32π + 32π
Simplifying:
64π
Therefore, the volume of the solid is 64π cubic units, which corresponds to answer choice b. 512π/3.
To find the volume of the solid, we need to integrate the area of each cross section as we move along the y-axis.
First, let's find the equation of the semicircle at any given y-value. Since the cross sections are perpendicular to the y-axis, the radius of each semicircle will equal the square root of the difference between the squared y-value and the constant term in the equation of the base circle.
The equation of the base circle is x^2 + y^2 = 16. By rearranging this equation, we can isolate x: x = sqrt(16 - y^2).
Now, the equation of the semicircle at any y-value becomes x = sqrt(16 - y^2). Note that the x-values will be positive on one side of the y-axis and negative on the other side.
To find the volume, we integrate the area of each semicircle from the lowest y-value to the highest y-value. The lowest y-value is -4, and the highest is 4 since the base circle has a radius of 4.
Let A(y) be the area of the cross section at y. The area of a semicircle is (1/2)πr^2, where r is the radius of the semicircle at that y-value.
A(y) = (1/2)π[(sqrt(16 - y^2))^2] = (1/2)π(16 - y^2)
To find the volume, we integrate A(y) from -4 to 4:
V = ∫[from -4 to 4] (1/2)π(16 - y^2) dy
V = (1/2)π ∫[from -4 to 4] (16 - y^2) dy
Integrating, we get:
V = (1/2)π [16y - (y^3)/3] |[-4 to 4]
V = (1/2)π [(16(4) - (4^3)/3) - (16(-4) - (-4^3)/3)]
V = (1/2)π [(64 - 256/3) - (-64 + 64/3)]
V = (1/2)π [(64/3) + (64/3)]
V = (1/2)π (128/3)
Therefore, the volume of the solid is 128π/3 cubic units.
Hence, the answer is (a) 128π/3.