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b) Using double integral, find the volume of the solid that lies under the para...
Oct 2, 2024
Solution by Steps
step 1
To find the volume under the paraboloid z=x2+y2z = x^2 + y^2 and above the xyxy-plane, we first need to set up the double integral. The region of integration is defined by the cylinder x2+y2=2xx^2 + y^2 = 2x. We can rewrite this as (x−1)2+y2=1(x - 1)^2 + y^2 = 1, which describes a circle of radius 1 centered at (1, 0)
step 2
We will convert to polar coordinates for easier integration. The transformation is given by x=rcos⁡θ+1x = r \cos \theta + 1 and y=rsin⁡θy = r \sin \theta. The Jacobian of this transformation is rr. The limits for rr will be from 0 to 1, and for θ\theta from 0 to 2π2\pi
step 3
The volume VV can be expressed as the double integral: V=∫02π∫01(r2cos⁡2θ+r2sin⁡2θ)r dr dθ=∫02π∫01r3 dr dθ. V = \int_0^{2\pi} \int_0^1 (r^2 \cos^2 \theta + r^2 \sin^2 \theta) r \, dr \, d\theta = \int_0^{2\pi} \int_0^1 r^3 \, dr \, d\theta.
step 4
Evaluating the inner integral: ∫01r3 dr=[r44]01=14. \int_0^1 r^3 \, dr = \left[ \frac{r^4}{4} \right]_0^1 = \frac{1}{4}. Now, evaluating the outer integral: V=∫02π14 dθ=14⋅2π=π2. V = \int_0^{2\pi} \frac{1}{4} \, d\theta = \frac{1}{4} \cdot 2\pi = \frac{\pi}{2}.
Answer
The volume of the solid is π2\frac{\pi}{2}.
Key Concept
The use of double integrals in polar coordinates to find volumes under surfaces.
Explanation
The volume is calculated by integrating the height of the surface over the area defined by the cylinder, using polar coordinates simplifies the integration process.
Solution by Steps
step 1
To evaluate the triple integral I=∭E16z dVI = \iiint_{E} 16z \, dV, we first need to express the region EE, which is the upper half of the sphere defined by x2+y2+z2=1x^2 + y^2 + z^2 = 1. In spherical coordinates, we have: x=ρsin⁡ϕcos⁡θx = \rho \sin \phi \cos \theta, y=ρsin⁡ϕsin⁡θy = \rho \sin \phi \sin \theta, and z=ρcos⁡ϕz = \rho \cos \phi, where ρ\rho ranges from 0 to 1, ϕ\phi ranges from 0 to π2\frac{\pi}{2} (upper half), and θ\theta ranges from 0 to 2π2\pi
step 2
The volume element in spherical coordinates is given by dV=ρ2sin⁡ϕ dρ dϕ dθdV = \rho^2 \sin \phi \, d\rho \, d\phi \, d\theta. Thus, we can rewrite the integral as: I=∫02π∫0π2∫0116(ρcos⁡ϕ)⋅ρ2sin⁡ϕ dρ dϕ dθ.I = \int_0^{2\pi} \int_0^{\frac{\pi}{2}} \int_0^1 16(\rho \cos \phi) \cdot \rho^2 \sin \phi \, d\rho \, d\phi \, d\theta.
step 3
We can factor out the constants and separate the integrals: I=16∫02πdθ∫0π2cos⁡ϕsin⁡ϕ dϕ∫01ρ3 dρ.I = 16 \int_0^{2\pi} d\theta \int_0^{\frac{\pi}{2}} \cos \phi \sin \phi \, d\phi \int_0^1 \rho^3 \, d\rho.
step 4
Evaluating the innermost integral: ∫01ρ3 dρ=[ρ44]01=14.\int_0^1 \rho^3 \, d\rho = \left[\frac{\rho^4}{4}\right]_0^1 = \frac{1}{4}. Now, for the middle integral: ∫0π2cos⁡ϕsin⁡ϕ dϕ=12∫0π2sin⁡(2ϕ) dϕ=12[−12cos⁡(2ϕ)]0π2=12.\int_0^{\frac{\pi}{2}} \cos \phi \sin \phi \, d\phi = \frac{1}{2} \int_0^{\frac{\pi}{2}} \sin(2\phi) \, d\phi = \frac{1}{2} \left[-\frac{1}{2} \cos(2\phi)\right]_0^{\frac{\pi}{2}} = \frac{1}{2}.
step 5
Finally, the outer integral: ∫02πdθ=2π.\int_0^{2\pi} d\theta = 2\pi. Combining all parts, we have: I=16⋅2π⋅12⋅14=16⋅π4=4π.I = 16 \cdot 2\pi \cdot \frac{1}{2} \cdot \frac{1}{4} = 16 \cdot \frac{\pi}{4} = 4\pi.
Answer
The value of the triple integral is 4π4\pi.
Key Concept
Evaluating triple integrals in spherical coordinates allows us to simplify the integration over three-dimensional regions, especially for symmetric shapes like spheres.
Explanation
The integral was evaluated by converting to spherical coordinates, simplifying the volume element, and separating the integrals, leading to the final result of 4π4\pi.
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