Section 9.6

Some Applications of Double Integrals in Mechanics and Strength of Materials

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Problem 3634

The coordinates of the center of gravity of a plate are found by the formulas:

Mxc=∬Sxμ dx dy,Myc=∬Syμ dx dy,M=∬Sμ dx dy, Mx_{c}=\iint \limits _{S} x\mu \, dx \, dy, \qquad My_{c}=\iint \limits _{S} y\mu \, dx \, dy, \qquad M=\iint \limits _{S} \mu \, dx \, dy,

where μ=μ(x,y)\mu =\mu (x,y) is the density of the plate at the point (x,y)(x,y); MM is the mass of the plate, and the quantities MxcMx_{c} and MycMy_{c} are the static moments of the plate relative to the axes OyOy and OxOx.

Find the coordinates of the center of gravity of a homogeneous plate bounded by the lines x=0x=0, y=0y=0, x+y=ax+y=a.

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Problem 3635

Find the coordinates of the center of gravity of a homogeneous plate bounded by the lines y2=2pxy^{2}=2px, y=0y=0, x=x0x=x_{0}.

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Problem 3636

Find the coordinates of the center of gravity of a homogeneous plate bounded by the line x2/3+y2/3=a2/3x^{2/3}+y^{2/3}=a^{2/3}; x>0x>0, y>0y>0.

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Problem 3637

Find the coordinates of the center of gravity of a homogeneous plate bounded by the line x2+y2=a2x^{2}+y^{2}=a^{2}; x>0x>0, y>0y>0.

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Problem 3638

Find the coordinates of the center of gravity of a homogeneous plate bounded by the line x2+y2=a2x^{2}+y^{2}=a^{2}; ∣y∣<tg⁡α\left|y\right|<\operatorname {tg}\alpha.

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Remark.
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In the source the condition is given literally as ∣y∣<tg⁡α\left|y\right|<\operatorname {tg}\alpha (with no factor of xx); the naturally expected condition for a sector is ∣y∣<xtg⁡α\left|y\right|<x\operatorname {tg}\alpha, which is consistent with the answer.

Problem 3639

Find the coordinates of the center of gravity of a homogeneous plate bounded by the line r2=a2cos⁡2φr^{2}=a^{2}\cos 2\varphi (the right-hand loop).

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Problem 3640

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

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Problem 3641

Find the coordinates of the center of gravity of a homogeneous plate bounded by the loop of the curve x3+y3=3axyx^{3}+y^{3}=3axy.

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Problem 3642

Find the coordinates of the center of gravity of a homogeneous plate bounded by the lines x=a(t−sin⁡t)x=a(t-\sin t), y=a(1−cos⁡t)y=a(1-\cos t); 0<t<2π0<t<2\pi; y=0y=0.

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Problem 3643

Find the coordinates of the center of gravity of a homogeneous plate bounded by the line x4+y4=x2yx^{4}+y^{4}=x^{2}y (the right-hand loop).

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Problem 3644

Find the coordinates of the center of gravity of a homogeneous plate bounded by the line x+y=a\sqrt{x}+\sqrt{y}=\sqrt{a}, x=0x=0, y=0y=0.

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Problem 3645

Find the coordinates of the center of gravity of a homogeneous plate bounded by the line (xa+yb)4=xyab\left(\dfrac {x}{a}+\dfrac {y}{b}\right)^{4}=\dfrac {xy}{ab} (the right-hand loop).

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Problem 3646

Prove that the volume of a right cylinder (or prism) cut off by a plane not necessarily parallel to the base equals the area of the base multiplied by the height above the center of gravity of the base.

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Problem 3647

Prove that the volume of the solid obtained by rotating an area SS about an axis that does not intersect SS and lies in the same plane equals the area SS multiplied by the length of the circle traced by the center of gravity of the area SS. (Pappus's theorem, often called Guldin's theorem.)

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Problem 3648

The moment of inertia of an area SS relative to some axis lying in the same plane is called the integral J=∬Sδ2 dx dy\displaystyle J=\iint \limits_{S} \delta^{2} \, dx \, dy, where δ\delta is the distance from the variable point (x,y)(x,y) to the axis, and the integral is taken over the whole area SS. In particular, the moments of inertia relative to the axes OxOx and OyOy equal the integrals Jx=∬Sy2 dx dyJ_{x}=\iint \limits_{S} y^{2} \, dx \, dy and Jy=∬Sx2 dx dyJ_{y}=\iint \limits_{S} x^{2} \, dx \, dy. The polar moment of the area SS relative to some point is called the integral ∬Sr2 dx dy\iint \limits_{S} r^{2} \, dx \, dy, where rr is the distance from the point (x,y)(x,y) to the given point. In particular, the polar moment relative to the origin equals ∬S(x2+y2) dx dy\iint \limits_{S}(x^{2}+y^{2}) \, dx \, dy. The centrifugal moment is called the integral Kxy=∬Sxy dx dyK_{xy}=\iint \limits_{S} xy \, dx \, dy. The curve Jxx2−2Kxyxy+Jyy2=1J_{x}x^{2}-2K_{xy}xy+J_{y}y^{2}=1 is called the ellipse of inertia of the area SS.

Find the moments of inertia of a sector with radius aa and central angle 2φ2\varphi relative to its axis of symmetry and relative to the perpendicular to it passing through the center of gravity.

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Problem 3649

Find the moment of inertia of a segment with radius aa and central angle 2φ2\varphi relative to its axis of symmetry.

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Problem 3650

Can the ellipse of inertia of a sector become a circle?

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Problem 3651

An isosceles triangle has base aa and height hh. For what relation between aa and hh does the ellipse of inertia of the triangle relative to the center of gravity become a circle?

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Problem 3652

Find the moments of inertia of the ellipse x2a2+y2b2=1\dfrac {x^{2}}{a^{2}}+\dfrac {y^{2}}{b^{2}}=1 relative to the axes, and find the equation of the ellipse of inertia.

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Problem 3653

A ring between concentric circles of radii 13 and 12 has the same area as a circle of radius 5. How many times greater is the moment of inertia of the ring than that of the circle, if both moments are taken relative to a diameter?

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Problem 3654

Find the moment of inertia relative to the axis OxOx of the parallelogram with sides a1x+b1y=±h1a_{1}x+b_{1}y=\pm h_{1}, a2x+b2y=±h2a_{2}x+b_{2}y=\pm h_{2}.

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Problem 3655

Find the centrifugal moment of inertia of the same parallelogram (with sides a1x+b1y=±h1a_{1}x+b_{1}y=\pm h_{1}, a2x+b2y=±h2a_{2}x+b_{2}y=\pm h_{2}).

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Problem 3656

Find the moment of inertia of the area of the curve x4+y4=x2+y2x^{4}+y^{4}=x^{2}+y^{2} relative to the axis OxOx and compare its value with the same moment of inertia of the square ∣x+y∣+∣x−y∣=2\left|x+y\right|+\left|x-y\right|=2.

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Problem 3657

The torsional rigidity of a rod with a simply connected cross-section SS is measured by the quantity C=2μ∬Sψ(x,y) dx dyC=2\mu \iint \limits_{S} \psi (x,y) \, dx \, dy, where μ\mu is the shear modulus of the rod's material, and ψ(x,y)\psi (x,y) is the stress function. The function ψ(x,y)\psi (x,y) equals zero on the boundary of the cross-section, and inside it satisfies the equation ∂2ψ∂x2+∂2ψ∂y2=−2\dfrac {\partial^{2}\psi }{\partial x^{2}}+\dfrac {\partial^{2}\psi }{\partial y^{2}}=-2.

Find the rigidity CC for a circular cross-section, when 2ψ=a2−x2−y22\psi =a^{2}-x^{2}-y^{2}.

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Problem 3658

Find CC for an elliptical cross-section, when (a2+b2)ψ=a2b2−b2x2−a2y2(a^{2}+b^{2})\psi =a^{2}b^{2}-b^{2}x^{2}-a^{2}y^{2}, where aa and bb are the semi-axes of the ellipse.

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Problem 3659

The same question (find the torsional rigidity CC) as in problem 3658, for an equilateral triangle with sides x±y3=hx\pm y\sqrt{3}=h, x=0x=0. In this case 2hψ=x(h−x−y3)(h−x+y3)2h\psi =x(h-x-y\sqrt{3})(h-x+y\sqrt{3}).

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Problem 3660

The level of a liquid is the axis OxOx, and the axis OyOy is directed downward. Prove that the pressure of the liquid on a vertical plate SS equals γ∬Sy dx dy\gamma \iint \limits_{S} y \, dx \, dy, where γ\gamma is the specific weight of the liquid. Prove that the point of application of the resultant of the pressure forces lies at a depth hh, where h∬Sy dx dy=∬Sy2 dx dy\displaystyle h\iint \limits_{S} y \, dx \, dy=\iint \limits_{S} y^{2} \, dx \, dy.

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