Flux across hemispheres and paraboloids Let S be the hemispherex2 + y2 + z2 = a2, for z ≥ 0, and let T be the paraboloid z = a - (x2 + y2)/a, for z ≥ 0, where a > 0. Assume the surfaces have outward normal vectors.a. Verify that S and T have the same base (x2 + y2 ≤ a2) and thesame high point (0, 0, a).b. Which surface has the greater area?c. Show that the flux of the radial field F = ⟨x, y, z⟩ across S is 2πa3.d. Show that the flux of the radial field F = ⟨x, y, z⟩ across T is 3πa3/2.

Answered: Flux across hemispheres and paraboloids…

Algebra & Trigonometry with Analytic Geometry

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ISBN:9781133382119

Author:Swokowski

Publisher:Swokowski

Chapter8: Applications Of Trigonometry

Section8.3: Vectors

Problem 60E

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Flux across hemispheres and paraboloids Let S be the hemisphere
x² + y² + z² = a², for z ≥ 0, and let T be the paraboloid z = a - (x² + y²)/a, for z ≥ 0, where a > 0. Assume the surfaces have outward normal vectors.
a. Verify that S and T have the same base (x² + y² ≤ a²) and the
same high point (0, 0, a).
b. Which surface has the greater area?
c. Show that the flux of the radial field F = ⟨x, y, z⟩ across S is 2πa³.
d. Show that the flux of the radial field F = ⟨x, y, z⟩ across T is 3πa³/2.

Expert Solution

Step 1

Since you have posted a question with multiple sub-parts, we will provide the solution only to the first three sub-parts as per our Q&A guidelines. Please repost the remaining sub-parts separately.

Given:

The equation of the hemisphere is $x^{2} + y^{2} + z^{2} = a^{2}$ , for $z \geq 0$ and

The equation of the paraboloid is $z = a - \frac{(x^{2} + y^{2})}{a}$ , for $z \geq 0$ , where $a > 0$ .

Flux across hemispheres and paraboloids:

The flux across a surface is a measure of the flow of a vector field through the surface. The flux across hemispheres and paraboloids can be found using the divergence theorem or by direct computation using surface integrals.

Flux across a hemisphere:

Consider a hemisphere with the radius R centred at the origin. The unit outward normal vector to the hemisphere is given by $n = \frac{(x, y, z)}{R}$ . The flux of a vector field F = (F1, F2, F3) across the hemisphere is given by:

flux $= \int \int F \cdot n d S$

Where the integral is taken over the surface of the hemisphere. Using spherical coordinates, the surface element can be expressed as $d S = R^{2} s i n θ d θ d φ$ , and the unit outward normal vector is given by $n = (\sin θ \cos φ, \sin θ \sin φ, \cos θ)$ . Thus, the flux can be shown as:

flux $= \int_{0}^{2 π} \int_{0}^{π} F \cdot n R^{2} s i n θ d θ d φ$

To evaluate the integral, we need to express the vector field F in terms of spherical coordinates. For example, if $F = (x^{2}, y^{2}, z^{2})$ , then $F \cdot n = \frac{(x^{3} + y^{3} + z^{3})}{R^{3}}$ . Substituting this expression in the above integral and evaluating the integrals gives the flux across the hemisphere.

Flux across a paraboloid:

Consider a paraboloid of revolution $z = x^{2} + y^{2}$ , bounded by a circle of radius R in the xy-plane. The unit outward normal vector to the paraboloid is given by $n = \frac{(2 x, 2 y, - 1)}{\sqrt{(1 + 4 x^{2} + 4 y^{2})}}$ . The flux of a vector field F = (F1, F2, F3) across the paraboloid is given by:

flux $= \int \int F \cdot n d S$

Where the integral is taken over the surface of the paraboloid. Using cylindrical coordinates, the surface element can be expressed as $d S = \sqrt{(1 + r^{2})} r d r d θ$ , and the unit outward normal vector is given by $n = \frac{(2 r \cos θ, 2 r \sin θ, - 1)}{\sqrt{(1 + 4 r^{2})}}$ . Thus, the flux can be shown as:

flux $= \int_{0}^{R} \int_{0}^{2 π} F \cdot n \sqrt{(1 + r^{2})} r d r d θ$

To evaluate the integral, we need to express the vector field F in terms of cylindrical coordinates. For example, suppose F = (x, y, z), then $F \cdot n = - \frac{z}{\sqrt{(1 + 4 r^{2})}}$ . Substituting this expression in the above integral and evaluating the integrals gives the flux across the paraboloid.

Step 2

a. To verify that S and T have the same base and high point, we can set z = a in the equation of the hemisphere and in the equation of the paraboloid and then solve for x and y. We have:

For the hemisphere:

$x^{2} + y^{2} + a^{2} = a^{2}$ , which simplifies to $x^{2} + y^{2} = 0$ . This means that the base of the hemisphere is the point (0, 0) and the high point is (0, 0, a).

For the paraboloid:

$z = a - \frac{(x^{2} + y^{2})}{a} = a - \frac{r^{2}}{a}$ , where $r^{2} = x^{2} + y^{2}$ . Setting z = a gives, which is the equation of a circle of radius a in the xy-plane. Therefore, the base of the paraboloid is the circle $x^{2} + y^{2} = a^{2}$ and the high point is (0, 0, a).

Thus, S and T have the same base (the circle $x^{2} + y^{2} = a^{2}$ ) and the same high point (0, 0, a).

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