Find the gradient vector of the function f that is defined by
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f(x, y, z) = x^{4} y^{2} - y^{3} z^{2} + z x .
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Response:
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(\nabla f)(x, y, z) = | | ( | 4 x^{3} y^{2} + z, 2 x^{4} y - 3 y^{2} z^{2}, -2 y^{3} z + x | ) |
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Find the divergence of the vector field F that is defined by
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F(x, y, z) = (z^{3} x, x^{3} y, y^{3} z) .
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Response:
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{\partial}/{\partial x}(z^{3} x) + {\partial}/{\partial y}(x^{3} y) + {\partial}/{\partial z}(y^{3} z) = z^{3} + x^{3} + y^{3} .
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Find the double integral of the function f(x, y) = x y^{2}
on the rectangle, having sides parallel to the coordinate axes, with
diagonally opposite vertices at the points (-1, 1) and (5, 3).
Response: The rectangle, say E, is described by the inequalities
Thus,
| INT[INT[_{E} x y^{2} dA ]] |
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| INT[_{-1}^{5} INT[_{1}^{3} x y^{2} dy ] dx ] |
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| INT[_{-1}^{5} x | | ( | INT[_{1}^{3} y^{2} dy ] | ) |
| dx ] |
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| INT[_{-1}^{5} x . |
| _{y = 1}^{y = 3} dx ] |
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| {26}/{3} INT[_{-1}^{5} x dx ] |
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| {26}/{3} |
| _{x = -1}^{x = 5} |
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| {26}/{3} | | ( | {5^{2}}/{2} - {(-1)^{2}}/{2} | ) |
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Find the tangent plane at the point (3, -1, 2) to the surface
Response: The chief task here is to find a vector normal to the tangent
plane to be used as coefficient vector in the equation for the plane.
Since the surface is a level set of the function
the gradient \nabla f of f at the given point must be normal to the
tangent plane at that point (unless it vanishes).
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| (y^{2} z, 2 x y z, x y^{2}) |
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Hence, the tangent plane has an equation of the form
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2 x - 12 y + 3 z = constant .
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Using the fact that (3, -1, 2) must satisfy this equation, one obtains
the equation
Find the arc length of the helix that is given parametrically by
for 0 <= t <= pi/2.
Response: This involves a straightforward application of the
definition of the length of a parameterized path as the integral of
the length of the derivative of a point on the path with respect
to the the parameter:
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| INT[_{0}^{pi/2} || | | ( | {dx}/{dt}, {dy}/{dt}, {dz}/{dt} | ) |
| || dt ] |
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| INT[_{0}^{pi/2} || |
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| INT[_{0}^{pi/2} SQRT{4^{2} |
| + 3^{2}} dt ] |
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Find the path integral of the vector field F that is defined by
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F(x, y, z) = (x - 2 z, y + z , 1 - y)
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over the path given by
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R(t) = (t^{3} , t , t^{2})
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for 0 <= t <= 2.
Response: This involves a straightforward application of the
definition of the integral over a parameterized path of a vector field:
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| INT[_{0}^{2} F |
| . R'(t) dt ] |
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| INT[_{0}^{2} | | ( | t^{3} - 2 t^{2}, t + t^{2}, 1 - t | ) |
| . |
| dt ] |
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| INT[_{0}^{2} | | ( | (3 t^{5} - 6 t^{4}) + (t^{2} + t) + (-2 t^{2} + 2 t) | ) |
| dt ] |
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| INT[_{0}^{2} (3 t^{5} - 6 t^{4} - t^{2} + 3 t) dt ] |
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| | [ | {t^{6}}/{2} - {6 t^{5}}/{5} - {t^{3}}/{3} + {3 t^{2}}/{2} | ] |
| _{t = 0}^{t = 2} |
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| 32 - {192}/{5} - {8}/{3} + 6 |
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Find the equation of the plane containing the point (-1, 3, 2)
that is normal to the line defined by the two equations
Response: A key fact for this problem is that a single linear
equation in 3 variables is the equation of a plane in space, and
the coefficient vector of such an equation is a (unique up to a
scalar multiple) normal vector to that plane.
Since a point in the required plane is given, one only needs to find a
vector normal to the plane, which is the same thing as a vector
parallel to the given line. The line is given as the intersection of
two planes, and any normal to either plane must be normal to a vector
parallel to their line of intersection. Thus a vector parallel to the
line may be obtained as the ``cross product'' of normals to
the two different planes, provided that vector is not 0 -- which
happens when and only when the two planes are parallel and, therefore,
do not determine a line of intersection.
One finds:
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(3, -4, 2) \times (9, -5, -3) = (22, 27, 21) .
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Hence, the required plane has an equation of the form
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22 x + 27 y + 21 z = constant ,
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and the constant is determined by the fact that the given point
must satisfy the equation:
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22 . -1 + 27 . 3 + 21 . 2 = -22 + 81 + 42 = 101 .
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Hence, the required plane has the equation:
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22 x + 27 y + 21 z = 101 .
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Find the centroid of the solid cone
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x^{2} + y^{2} <= z^{2} , 0 <= z <= 2 .
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Response: The lateral boundary of the solid cone,
given by x^{2} + y^{2} = z^{2}, is the surface obtained by rotating
the line z = y in the plane x = 0 about the z-axis. Thus,
for z >= 0 the section of the solid cone by the plane normal to
the z-axis through the point (0, 0, z) is the disk of radius
z with center (0, 0, z). The radius of this (inverted) cone's
base, as well as its altitude, is 2, and, therefore its volume is
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{1}/{3} pi a^{2} h = {8pi}/{3} .
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By symmetry its centroid lies on the z-axis, and the only issue is
what is its z-coordinate:
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| {1}/{volume} INT[INT[INT[ z dV]]] |
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| {3}/{8pi} INT[_{0}^{2} z dz INT[INT[_{x^{2} + y^{2} <= z^{2}} dx dy]]] |
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| {3}/{8pi} INT[_{0}^{2} z . | | ( | Area of disk of radius z | ) |
| dz ] |
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| {3}/{8pi} INT[_{0}^{2} pi z^{3} dz ] |
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Hence, the centroid of the cone is the point (0, 0, 3pi/2).
Find the surface integral of the vector field F over the sphere,
oriented by its outer normal, of radius 3 with center at the origin
when F is given by
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F(x, y, z) = (x^{3} , y^{3} , z^{3}) .
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Response: The vector field is well-behaved everywhere, particularly
in the ball surrounded by the given sphere. The divergence theorem may
be applied.
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(div F)(x, y, z) = ({\partial}/{\partial x}, {\partial}/{\partial y}, {\partial}/{\partial z}). F = {\partial}/{\partial x}(x^{3}) + {\partial}/{\partial y}(y^{3}) + {\partial}/{\partial z}(z^{3}) = 3(x^{2} + y^{2} + z^{2}) .
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Therefore,
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INT[INT[_{S} F . N d sigma ]] = INT[INT[INT[_{B} div F d V ]]] .
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In polar coordinates, observing that
d V = rho^{2} sin phi drho dphi dtheta and that
div F = 3 rho^{2}, the triple integral with radius a = 3 becomes:
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3 INT[_{0}^{2pi} dtheta INT[_{0}^{pi} sin phi dphi INT[_{0}^{3} rho^{4} drho ]]] ,
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which evaluated becomes
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{12 pi}/{5} a^{5} = {2916 pi}/{5} .
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For a conservative vector field in the plane what can be said
about its integral over the circle, traversed counterclockwise, of radius
2 with center at the point (3, 4) ? Explain your answer.
Response: Because conservative in this context means that
the vector field is the gradient of a scalar, one may use the formula
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INT[_{C} \nabla f] = f(B) - f(A)
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where A and B are, respectively, the initial and final points of
an oriented curve C. A trip around a circle, no matter where it
begins, will end where it begins. Thus, A = B, and the integral
of a vector field that is conservative in the plane around any
circle must be zero.