Let V be the volume of the three-dimensional structure bounded by the region 0≤z≤1−x^2−y^2. If V=a/bπ, where a and b are positive coprime integers, what is a+b?
One easy way to do this is to consider the figure to be a solid of revolution formed by revolving the parabola z = 1-x^2 around the z-axis.
v = ∫[0,1] πr^2 dz
where r = x
v = π∫[0,1] (1-z) dz
= π/2
The other, more direct way, of course, is to do a volume integral:
v = 4∫[0,1]∫[0,√(1-x^2)]∫[0,1-x^2-y^2] dz dy dx
= 4∫[0,1]∫[0,√(1-x^2)] 1-x^2-y^2 dy dx
= 4∫[0,1] (1-x^2)y - 1/3 y^3 [0,√(1-x^2)] dx
= 4∫[0,1] (1-x^2)^(3/2) - 1/3 (1-x^2)^(3/2) dx
= 8/3∫[0,1] (1-x^2)^(3/2) dx
= 8/3 * 1/8 (x√(1-x^2) (5-2x^2) + 3arcsin(x)) [0,1]
= π/2
thank you steve!
Another easy way is to use cylindrical coordinates, like polar coordinates with a z-axis.
Just as dA = r dr dθ is the area element in polar coordinates,
dV = r dz dr dθ in cylindrical coordinates.
So, now we have z = 1-r^2
v = ∫[0,1]∫[0,2π]∫[0,1-r^2] r dz dr dθ
= ∫[0,1]∫[0,2π] r(1-r^2) dr dθ
= ∫[0,1] 2πr(1-r^2) dr
= 2π(1/2 r^2 - 1/4 r^4)
= π/2
To calculate the volume of the three-dimensional structure bounded by the region 0 ≤ z ≤ 1 − x^2 − y^2, we need to integrate the given region over the entire space.
The given region can be visualized as a parabolic shape that is vertically stretched along the z-axis. We can think of it as a solid that is formed by stacking a series of infinitesimally thin disks with varying radii.
To find the volume, we need to perform a triple integral over the region. Let's set up the integral.
∫∫∫ (1 − x^2 − y^2) dz dy dx
The limits of integration can be determined by examining the region. Here, the region is bounded by the paraboloid z = 1 − x^2 − y^2, with the z-axis as the lower bound. Since the paraboloid extends to infinity, the limits for x and y are also from negative infinity to positive infinity.
∫∫∫ (1 − x^2 − y^2) dz dy dx
= ∫∫ (1 − x^2 − y^2) dy dx
= ∫ (1 − x^2 − y^2) dx
To evaluate this integral, we can apply the formula for integrating polynomials:
∫ (1 − x^2 − y^2) dx = x − (x^3/3) − xy^2 + C
Now, we need to evaluate this expression by substituting appropriate limits of integration.
∫∫∫ (1 − x^2 − y^2) dz dy dx
= ∫∫ (x − (x^3/3) − xy^2 + C) dy dx
To solve this integral, we can integrate with respect to y and then with respect to x. However, since the question asks for the volume in terms of a fraction with π, there might be a simplification or symmetry that allows us to solve the integral without further calculations.
Unfortunately, without any additional information or simplifications, we cannot determine the specific value of the volume V or the sum of a and b.