Chapter 14
Calculus Early Transcendentals: Pearson New International Edition · 124 exercises
Problem 1
In Problems 1-6, use Stokes's Theorem to calculate $$ \iint_{S}(\operatorname{curl} \mathbf{F}) \cdot \mathbf{n} d S $$ \(\mathbf{F}=x^{2} \mathbf{i}+y^{2} \mathbf{j}+z^{2} \mathbf{k} ; \quad S\) is the hemisphere \(z=\) \(\sqrt{1-x^{2}-y^{2}}\) and \(\mathbf{n}\) is the upper normal.
4 step solution
Problem 1
In Problems 1-6, sketch a sample of vectors for the given vector field \(\mathbf{F}\). $$ \mathbf{F}(x, y)=x \mathbf{i}+y \mathbf{j} $$
5 step solution
Problem 1
In Problems 1-14, use Gauss's Divergence Theorem to calculate \(\iint_{\partial S} \mathbf{F} \cdot \mathbf{n} d S .\) \(\mathbf{F}(x, y, z)=z \mathbf{i}+x \mathbf{j}+y \mathbf{k} ; S\) is the hemisphere \(0 \leq z \leq \sqrt{9-x^{2}-y^{2}}\)
7 step solution
Problem 1
Use Green's Theorem to evaluate the given line integral. Begin by sketching the region \(S\). \(\oint_{C} 2 x y d x+y^{2} d y\), where \(C\) is the closed curve formed by \(y=x / 2\) and \(y=\sqrt{x}\) between \((0,0)\) and \((4,2)\)
7 step solution
Problem 1
$$ \int_{C}\left(x^{3}+y\right) d s ; C \text { is the curve } x=3 t, y=t^{3}, 0 \leq t \leq 1 $$
7 step solution
Problem 1
Evaluate \(\iint g(x, y, z) d S\). \(g(x, y, z)=x^{2}+y^{2}+z ; G: z=x+y+1,0 \leq x \leq 1\), \(0 \leq y \leq 1\)
8 step solution
Problem 1
Evaluate each line integral. \(\int_{C}\left(x^{3}+y\right) d s ; C\) is the curve \(x=3 t, y=t^{3}, 0 \leq t \leq 1\).
7 step solution
Problem 2
In Problems 1-6, sketch a sample of vectors for the given vector field \(\mathbf{F}\). $$ \mathbf{F}(x, y)=x \mathbf{i}-y \mathbf{j} $$
5 step solution
Problem 2
In Problems 1-14, use Gauss's Divergence Theorem to calculate \(\iint_{\partial S} \mathbf{F} \cdot \mathbf{n} d S .\) \(\mathbf{F}(x, y, z)=x \mathbf{i}+2 y \mathbf{j}+3 z \mathbf{k} ; S\) is the cube \(0 \leq x \leq 1,0 \leq y \leq 1,0 \leq z \leq 1\).
5 step solution
Problem 2
Use Green's Theorem to evaluate the given line integral. Begin by sketching the region \(S\). \(\oint_{C} \sqrt{y} d x+\sqrt{x} d y\), where \(C\) is the closed curve formed by \(y=0, x=2\), and \(y=x^{2} / 2\)
6 step solution
Problem 2
$$ \int_{C} x y^{2 / 5} d s ; C \text { is the curve } x=\frac{1}{2} t, y=t^{5 / 2}, 0 \leq t \leq 1 \text {. } $$
6 step solution
Problem 2
Evaluate each line integral. \(\int_{C} x y^{2 / 5} d s ; C\) is the curve \(x=\frac{1}{2} t, y=t^{5 / 2}, 0 \leq t \leq 1\).
6 step solution
Problem 3
In Problems 1-6, use Stokes's Theorem to calculate \(\mathbf{F}=(y+z) \mathbf{i}+\left(x^{2}+z^{2}\right) \mathbf{j}+y \mathbf{k} ; S\) is the half-cylinder \(z=\sqrt{1-x^{2}}\) between \(y=0\) and \(y=1\) and \(\mathbf{n}\) is the upper normal.
7 step solution
Problem 3
In Problems 1-6, sketch a sample of vectors for the given vector field \(\mathbf{F}\). $$ \mathbf{F}(x, y)=-x \mathbf{i}+2 y \mathbf{j} $$
5 step solution
Problem 3
In Problems 1-14, use Gauss's Divergence Theorem to calculate \(\iint_{\partial S} \mathbf{F} \cdot \mathbf{n} d S .\) \(\mathbf{F}(x, y, z)=\cos z^{2} \mathbf{i}+y \mathbf{j}+\cos x^{2} \mathbf{k} ; S\) is the cube \(-1 \leq x \leq 1,-1 \leq y \leq 1,-1 \leq z \leq 1\).
5 step solution
Problem 3
Use Green's Theorem to evaluate the given line integral. Begin by sketching the region \(S\). \(\oint_{C}\left(2 x+y^{2}\right) d x+\left(x^{2}+2 y\right) d y\), where \(C\) is the closed curve formed by \(y=0, x=2\), and \(y=x^{3} / 4\)
7 step solution
Problem 3
\(\int_{C}(\sin x+\cos y) d s ; C\) is the line segment from \((0,0)\) to \((\pi, 2 \pi)\).
4 step solution
Problem 3
Evaluate each line integral. \(\int_{C}(\sin x+\cos y) d s ; C\) is the line segment from \((0,0)\) to \((\pi, 2 \pi)\).
5 step solution
Problem 4
In Problems 1-6, use Stokes's Theorem to calculate \(\mathbf{F}=x z^{2} \mathbf{i}+x^{3} \mathbf{j}+\cos x z \mathbf{k} ; S\) is the part of the ellipsoid \(x^{2}+y^{2}+3 z^{2}=1\) below the \(x y\)-plane and \(\mathbf{n}\) is the lower normal.
6 step solution
Problem 4
In Problems 1-6, sketch a sample of vectors for the given vector field \(\mathbf{F}\). $$ \mathbf{F}(x, y)=3 x \mathbf{i}+y \mathbf{j} $$
4 step solution
Problem 4
Use Green's Theorem to evaluate the given line integral. Begin by sketching the region \(S\). \(\oint_{C} x y d x+(x+y) d y\), where \(C\) is the triangle with vertices \((0,0),(2,0)\), and \((0,1)\)
7 step solution
Problem 4
$$ \int_{C} x e^{y} d s ; C \text { is the line segment from }(-1,2) \text { to }(1,1) \text {. } $$
5 step solution
Problem 4
Evaluate \(\iint g(x, y, z) d S\). \(g(x, y, z)=2 y^{2}+z ; G: z=x^{2}-y^{2}, 0 \leq x^{2}+y^{2} \leq 1\)
7 step solution
Problem 4
Evaluate each line integral. \(\int_{C} x e^{y} d s ; C\) is the line segment from \((-1,2)\) to \((1,1)\).
5 step solution
Problem 5
In Problems 1-6, sketch a sample of vectors for the given vector field \(\mathbf{F}\). $$ \mathbf{F}(x, y, z)=x \mathbf{i}+0 \mathbf{i}+\mathbf{k} $$
4 step solution
Problem 5
In Problems 1-14, use Gauss's Divergence Theorem to calculate \(\iint_{\partial S} \mathbf{F} \cdot \mathbf{n} d S .\) \(\mathbf{F}(x, y, z)=x^{2} y z \mathbf{i}+x y^{2} z \mathbf{j}+x y z^{2} \mathbf{k} ; S\) is the box \(0 \leq x \leq a, 0 \leq y \leq b, 0 \leq z \leq c\).
7 step solution
Problem 5
Use Green's Theorem to evaluate the given line integral. Begin by sketching the region \(S\). \(\oint_{C}\left(x^{2}+4 x y\right) d x+\left(2 x^{2}+3 y\right) d y\), where \(C\) is the ellipse \(9 x^{2}+16 y^{2}=144\)
5 step solution
Problem 5
\(\int_{C}(2 x+9 z) d s ; C\) is the curve \(x=t, y=t^{2}, z=t^{3}\), \(0 \leq t \leq 1\).
4 step solution
Problem 5
Evaluate each line integral. \(\int_{C}(2 x+9 z) d s ; C\) is the curve \(x=t, y=t^{2}, z=t^{3}\), \(0 \leq t \leq 1\).
6 step solution
Problem 6
In Problems 1-6, use Stokes's Theorem to calculate \(\mathbf{F}=(z-y) \mathbf{i}+(z+x) \mathbf{j}-(x+y) \mathbf{k} ; S\) is the part of the paraboloid \(z=1-x^{2}-y^{2}\) above the \(x y\)-plane and \(\mathbf{n}\) is the upward normal.
5 step solution
Problem 6
In Problems 1-14, use Gauss's Divergence Theorem to calculate \(\iint_{\partial S} \mathbf{F} \cdot \mathbf{n} d S .\) \(\mathbf{F}(x, y, z)=3 x \mathbf{i}-2 y \mathbf{j}+4 z \mathbf{k} ; S\) is the solid sphere \(x^{2}+y^{2}+z^{2} \leq 9 .\)
4 step solution
Problem 6
Use Green's Theorem to evaluate the given line integral. Begin by sketching the region \(S\). \(\oint_{C}\left(e^{3 x}+2 y\right) d x+\left(x^{2}+\sin y\right) d y\), where \(C\) is the rectangle with vertices \((2,1),(6,1),(6,4)\), and \((2,4)\).
5 step solution
Problem 6
\(\int_{C}\left(x^{2}+y^{2}+z^{2}\right) d s ; \quad C\) is the curve \(x=4 \cos t\), \(y=4 \sin t, z=3 t, 0 \leq t \leq 2 \pi\).
6 step solution
Problem 6
Evaluate each line integral. \(\int_{C}\left(x^{2}+y^{2}+z^{2}\right) d s ; \quad C\) is the curve \(x=4 \cos t\), \(y=4 \sin t, z=3 t, 0 \leq t \leq 2 \pi .\)
7 step solution
Problem 7
In Problems \(7-12\), use Stokes's Theorem to calculate \(\oint_{C} \mathbf{F} \cdot \mathbf{T} d s\). \(\mathbf{F}=2 z \mathbf{i}+x \mathbf{j}+3 y \mathbf{k} ; C\) is the ellipse that is the intersection of the plane \(z=x\) and the cylinder \(x^{2}+y^{2}=4\), oriented clockwise as viewed from above.
7 step solution
Problem 7
In Problems 7-12, find \(\nabla f\). $$ f(x, y, z)=x^{2}-3 x y+2 z $$
5 step solution
Problem 7
In Problems 1-14, use Gauss's Divergence Theorem to calculate \(\iint_{\partial S} \mathbf{F} \cdot \mathbf{n} d S .\) \(\mathbf{F}(x, y, z)=x^{2} \mathbf{i}+y^{2} \mathbf{j}+z^{2} \mathbf{k} ; S\) is the parabolic solid \(0 \leq z \leq 4-x^{2}-y^{2} .\)
6 step solution
Problem 7
\(\int_{C} y d x+x^{2} d y\); \(C\) is the curve \(x=2 t, \quad y=t^{2}-1\), \(0 \leq t \leq 2\).
5 step solution
Problem 7
Evaluate each line integral. \(\int_{C} y d x+x^{2} d y ; C\) is the curve \(x=2 t, \quad y=t^{2}-1\), \(0 \leq t \leq 2\).
8 step solution
Problem 8
In Problems \(7-12\), use Stokes's Theorem to calculate \(\oint_{C} \mathbf{F} \cdot \mathbf{T} d s\). \(\mathbf{F}=y \mathbf{i}+z \mathbf{j}+x \mathbf{k} ; C\) is the triangular curve with vertices \((0,0,0),(2,0,0)\), and \((0,2,2)\), oriented counterclockwise as viewed from above.
5 step solution
Problem 8
In Problems 7-12, find \(\nabla f\). $$ f(x, y, z)=\sin (x y z) $$
5 step solution
Problem 8
In Problems 1-14, use Gauss's Divergence Theorem to calculate \(\iint_{\partial S} \mathbf{F} \cdot \mathbf{n} d S .\) \(\mathbf{F}(x, y, z)=\left(x^{2}+\cos y z\right) \mathbf{i}+\left(y-e^{z}\right) \mathbf{j}+\left(z^{2}+x^{2}\right) \mathbf{k} ; S\) is the solid bounded by \(x^{2}+y^{2}=4, x+z=2, z=0\).
8 step solution
Problem 8
\(\int_{C} y d x+x^{2} d y\), \(C\) is the right-angle curve from \((0,-1)\) to \((4,-1)\) to \((4,3)\).
5 step solution
Problem 8
Evaluate \(\iint g(x, y, z) d S\). \(g(x, y, z)=z ; G\) is the tetrahedron bounded by the coordinate planes and the plane \(4 x+8 y+2 z=16\).
9 step solution
Problem 8
Evaluate each line integral. \(\int_{C} y d x+x^{2} d y ; C\) is the right-angle curve from \((0,-1)\) to \((4,-1)\) to \((4,3)\).
5 step solution
Problem 9
In Problems \(7-12\), use Stokes's Theorem to calculate \(\oint_{C} \mathbf{F} \cdot \mathbf{T} d s\). \(\mathbf{F}=(y-x) \mathbf{i}+(x-z) \mathbf{j}+(x-y) \mathbf{k} ; C\) is the boundary of the plane \(x+2 y+z=2\) in the first octant, oriented clockwise as viewed from above.
5 step solution
Problem 9
In Problems 7-12, find \(\nabla f\). $$ f(x, y, z)=\ln |x y z| $$
5 step solution
Problem 9
In Problems 1-14, use Gauss's Divergence Theorem to calculate \(\iint_{\partial S} \mathbf{F} \cdot \mathbf{n} d S .\) \(\mathbf{F}(x, y, z)=\left(x+z^{2}\right) \mathbf{i}+\left(y-z^{2}\right) \mathbf{j}+x \mathbf{k} ; S\) is the solid \(0 \leq y^{2}+z^{2} \leq 1,0 \leq x \leq 2\)
7 step solution
Problem 9
Use the vector forms of Green's Theorem to calculate (a) \(\oint_{C} \mathbf{F} \cdot \mathbf{n} d s\) and (b) \(\oint_{C} \mathbf{F} \cdot \mathbf{T} d s\). \(\mathbf{F}=y^{2} \mathbf{i}+x^{2} \mathbf{j}\); \(C\) is the boundary of unit square with vertices \((0,0),(1,0),(1,1)\), and \((0,1)\).
10 step solution
Problem 9
\(\int_{C} y^{3} d x+x^{3} d y ; C\) is the right-angle curve from \((-4,1)\) to \((-4,-2)\) to \((2,-2)\).
6 step solution