Section 9.5

Line Integrals

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Problem 3617

If the equation of a curve is given either as a single equation f(x,y)=0f(x,y)=0, or parametrically: x=φ(t)x=\varphi (t), y=ψ(t)y=\psi (t), then the line integral along the curve ABAB, denoted by ∫ABP(x,y) dx+Q(x,y) dy\displaystyle \int_{AB} P(x,y) \, dx+Q(x,y) \, dy, is evaluated by reducing it to an ordinary definite integral: if x=φ(t)x=\varphi (t), y=ψ(t)y=\psi (t) and the parameter tt varies from aa to bb as we pass from AA to BB, then

∫ABP dx+Q dy=∫ab(P(φ,ψ)φ′+Q(φ,ψ)ψ′) dt. \int _{AB} P \, dx+Q \, dy=\int _{a}^{b} \big(P(\varphi ,\psi )\varphi '+Q(\varphi ,\psi )\psi '\big) \, dt.

When the direction of the path of integration is reversed, the value of the integral is multiplied by minus one: ∫ABP dx+Q dy=−∫BAP dx+Q dy\displaystyle \int_{AB} P \, dx+Q \, dy=-\int_{BA} P \, dx+Q \, dy.

Find the value of the integral ∫ABx dy−y dx\displaystyle \int_{AB} x \, dy-y \, dx along the parabola y=x2y=x^{2} between the points A(0,0)A(0,0) and B(2,4)B(2,4).

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Problem 3618

Find the integral ∫AB(x+y) dx+(x−y) dy\displaystyle \int_{AB}(x+y) \, dx+(x-y) \, dy, if the path of integration is the parabola y=x2y=x^{2} between the points (−1,1)(-1,1), (1,1)(1,1).

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Problem 3619

Find the integral ∫ABx2 ds\displaystyle \int_{AB} x^{2} \, ds, where the path of integration is the upper half of the circle x2+y2=a2x^{2}+y^{2}=a^{2} between the points A(a,0)A(a,0) and B(−a,0)B(-a,0).

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Problem 3620

Find the value of the integral ∫ABx dy\displaystyle \int_{AB} x \, dy, taken along the right half-circle x2+y2=a2x^{2}+y^{2}=a^{2} between the points A(0,−a)A(0,-a) and B(0,a)B(0,a).

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Problem 3621

Find the value of the integral ∫AB(x2+y2) dx+(x2−y2) dy\displaystyle \int_{AB}(x^{2}+y^{2}) \, dx+(x^{2}-y^{2}) \, dy, taken around the contour of the triangle with vertices at the points A(0,0)A(0,0), B(1,0)B(1,0), C(0,1)C(0,1).

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Remark.
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Green's formula: ∫CP dx+Q dy=∬S(∂Q∂x−∂P∂y)dx dy\displaystyle \int_{C} P \, dx+Q \, dy=\iint \limits_{S}\left(\dfrac {\partial Q}{\partial x}-\dfrac {\partial P}{\partial y}\right) dx \, dy transforms the line integral taken around a closed contour CC in the positive direction of traversal (i.e., counterclockwise) into an integral over the area SS bounded by that contour. It follows that the integral ∫ABP dx+Q dy\int_{AB} P \, dx+Q \, dy, taken between the points AA and BB along two different paths, has the same value if, in the region enclosed between these paths, the functions PP and QQ satisfy ∂Q∂x=∂P∂y\dfrac {\partial Q}{\partial x}=\dfrac {\partial P}{\partial y}.

Problem 3622

From Green's formula (see problem 3621) derive the equality: S=12∫C(x dy−y dx)S=\dfrac {1}{2}\int_{C}(x \, dy-y \, dx), where CC is a contour enclosing the area SS in the positive direction.

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Problem 3623

Using the formula of problem 3622 (S=12∫C(x dy−y dx)S=\dfrac {1}{2}\int_{C}(x\, dy-y\, dx)), compute the area of the ellipse x=acos⁡tx=a\cos t, y=bsin⁡ty=b\sin t.

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Problem 3624

Using the formula of problem 3622 (S=12∫C(x dy−y dx)S=\dfrac {1}{2}\int_{C}(x\, dy-y\, dx)), compute the area of the sector of the hyperbola x=ach⁡tx=a\operatorname {ch}t, y=bsh⁡ty=b\operatorname {sh}t, bounded by the axis OxOx, the arc of the hyperbola up to the point M(x0,y0)M(x_{0},y_{0}), and the line OMOM.

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Problem 3625

Using the formula of problem 3622 (S=12∫C(x dy−y dx)S=\dfrac {1}{2}\int_{C}(x\, dy-y\, dx)), compute the area of the sector of the astroid x=acos⁡3tx=a\cos^{3}t, y=asin⁡3ty=a\sin^{3}t, bounded by the lines OM1OM_{1} and OM2OM_{2} from the origin and the arc M1M2M_{1}M_{2} of the astroid (the points M1M_{1}, M2M_{2} correspond to the parameter values t1t_{1}, t2t_{2}).

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Problem 3626

Find the area of the three-cusped hypocycloid:

x=a(2cos⁡t+cos⁡2t),y=a(2sin⁡t−sin⁡2t). x=a(2\cos t+\cos 2t), \qquad y=a(2\sin t-\sin 2t).
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Problem 3627

Find the value of the line integral ∫Cx dy−y dxx2+y2\displaystyle \int_{C} \dfrac {x \, dy-y \, dx}{x^{2}+y^{2}}, taken around a closed contour CC in the positive direction.

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Problem 3628

The curves X(x,y)=0X(x,y)=0 and Y(x,y)=0Y(x,y)=0 have several simple points of intersection. Compute the value of the integral 12π∫CX dY−Y dXX2+Y2\dfrac {1}{2\pi }\displaystyle \int_{C} \dfrac {X\, dY-Y\, dX}{X^{2}+Y^{2}}, taken in the positive direction around a closed contour CC.

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Problem 3629

One application of line integrals is finding the center of gravity (xc,yc,zc)(x_{c}, y_{c}, z_{c}) of a curve. For this the following formulas are used:

Mxc=∫ABxμ ds,Myc=∫AByμ ds,Mzc=∫ABzμ ds,M=∫ABμ ds, Mx_{c}=\int _{AB} x\mu \, ds, \qquad My_{c}=\int _{AB} y\mu \, ds, \qquad Mz_{c}=\int _{AB} z\mu \, ds, \qquad M=\int _{AB} \mu \, ds,

where μ(x,y,z)\mu (x,y,z) is the density of the curve at the point (x,y,z)(x,y,z), and MM is the mass of the curve.

Find the coordinates of the center of gravity of a homogeneous arc of a circle of radius aa subtending a central angle 2φ2\varphi.

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Problem 3630

Find the coordinates of the center of gravity of the entire arc of the homogeneous cardioid r=a(1+cos⁡φ)r=a(1+\cos \varphi ).

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Problem 3631

Find the coordinates of the center of gravity of one arch of the homogeneous cycloid

x=a(t−sin⁡t),y=a(1−cos⁡t)for 0<t<2π. x=a(t-\sin t), \qquad y=a(1-\cos t) \qquad \text{for } 0<t<2\pi .
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Problem 3632

Find the coordinates of the center of gravity of an arc of the helix

x=acos⁡t,y=asin⁡t,z=ht x=a\cos t, \qquad y=a\sin t, \qquad z=ht

for 0<t<m0<t<m.

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Problem 3633

Find the coordinates of the center of gravity of a homogeneous arc of the curve

x=e−tcos⁡t,y=e−tsin⁡t,z=e−t x=e^{-t}\cos t, \qquad y=e^{-t}\sin t, \qquad z=e^{-t}

for 0<t<∞0<t<\infty.

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