Domains of Functions of Several Variables

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Determine and sketch the domain of existence of the following function: u=x+yu=x+\sqrt{y} .

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Determine and sketch the domain of existence of the following function: u=1−x2+y2−1u=\sqrt{1-x^{2}}+\sqrt{y^{2}-1} .

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Determine and sketch the domain of existence of the following function: u=1−x2−y2u=\sqrt{1-x^{2}-y^{2}} .

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Determine and sketch the domain of existence of the following function: u=1x2+y2−1u=\dfrac {1}{\sqrt{x^{2}+y^{2}-1}}.

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Determine and sketch the domain of existence of the following function: u=(x2+y2−1)(4−x2−y2)u=\sqrt{(x^{2}+y^{2}-1)(4-x^{2}-y^{2})}.

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Determine and sketch the domain of existence of the following function: u=x2+y2−x2x−x2−y2u=\sqrt{\dfrac {x^{2}+y^{2}-x}{2x-x^{2}-y^{2}}}.

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Determine and sketch the domain of existence of the following function: u=1−(x2+y)2u=\sqrt{1-(x^{2}+y)^{2}}.

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Determine and sketch the domain of existence of the following function: u=ln⁡(−x−y)u=\ln (-x-y).

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Determine and sketch the domain of existence of the following function: u=arcsin⁡yxu=\arcsin \dfrac {y}{x}.

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Determine and sketch the domain of existence of the following function: u=arccos⁡xx+yu=\arccos \dfrac {x}{x+y}.

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Determine and sketch the domain of existence of the following function: u=arcsin⁡xy2+arcsin⁡(1−y)u=\arcsin \dfrac {x}{y^{2}}+\arcsin (1-y).

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Determine and sketch the domain of existence of the following function: u=sin⁡(x2+y2)u=\sqrt{\sin (x^{2}+y^{2})}.

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Determine and sketch the domain of existence of the following function: u=arccos⁡zx2+y2u=\arccos \dfrac {z}{\sqrt{x^{2}+y^{2}}}.

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Determine and sketch the domain of existence of the following function: u=ln⁡(xyz)u=\ln (xyz).

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Determine and sketch the domain of existence of the following function: u=ln⁡(−1−x2−y2+z2)u=\ln (-1-x^{2}-y^{2}+z^{2}).

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Problem 3.2.8 from Кудрявцев Л.Д., Кутасов А.Д., Чехлов В.И., Шабунин М.И. "Функции многих переменных" [ККЧ+03]More problems in Subsection 3.2.2 "Problems"

Find the domain of the function of two variables given by the formula:

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(1)

u=x+y+x−yu = \sqrt{x+y}+\sqrt{x-y};

(2)

u=1−x2−y2u = \sqrt{1-x^{2}-y^{2}};

(3)

u=1x2+y2−1u = \frac{1}{\sqrt{x^{2}+y^{2}-1}};

(4)

u=ln⁡(x2+y2−1)u = \ln \left(x^{2}+y^{2}-1\right);

(5)

u=ln⁡(y2−4x+8)u = \ln \left(y^{2}-4 x+8\right);

(6)

u=ln⁡(x2+4y2−2x−3)u = \ln \left(x^{2}+4 y^{2}-2 x-3\right);

(7)

u=1x2−16(y2+1)u = \frac{1}{\sqrt{x^{2}-16\left(y^{2}+1\right)}};

(8)

u=ln⁡x2+y2−x2x−x2−y2u = \ln \frac{x^{2}+y^{2}-x}{2 x-x^{2}-y^{2}};

(9)

u=1−∣x∣−∣y∣u = \sqrt{1-\left|x\right|-\left|y\right|};

(10)

u=15x2−4xy+2y2+2x−2y+3u = \frac{1}{5 x^{2}-4 x y+2 y^{2}+2 x-2 y+3};

(11)

u=ln⁡xln⁡y1−x−yu = \frac{\ln x \ln y}{\sqrt{1-x-y}};

(12)

u=4x2−x3−4xu = \sqrt{4 x^{2}-x^{3}-4 x};

(13)

u=log⁡a(2−x2−y2)u = \sqrt{\log_{a}\left(2-x^{2}-y^{2}\right)};

(14)

u=(x2+y2−a)(2a−x2−y2)u = \sqrt{\left(x^{2}+y^{2}-a\right)\left(2 a-x^{2}-y^{2}\right)};

(15)

u=ln⁡(3x+y−3)+ln⁡(3−x)3x−2y+6u = \ln (3 x+y-3)+\frac{\ln (3-x)}{\sqrt{3 x-2 y+6}};

(16)

u=x−1+y−1+x+y−3+42−6x−7yu = \sqrt{x-1}+\sqrt{y-1}+\sqrt{x+y-3}+\sqrt{42-6 x-7 y};

(17)

u=ysin⁡xu = \sqrt{y \sin x};

(18)

u=ln⁡x−ln⁡sin⁡yu = \ln x-\ln \sin y;

(19)

u=ln⁡sin⁡π(x2+y2)u = \ln \sin \pi \left(x^{2}+y^{2}\right);

(20)

u=xln⁡tg⁡(πyx)u = \sqrt{x} \ln \operatorname {tg}\left(\frac{\pi y}{x}\right);

(21)

u=arccos⁡yu = \arccos y;

(22)

u=arccos⁡(x+y)u = \arccos (x+y);

(23)

u=arcsin⁡xx+yu = \arcsin \frac{x}{x+y};

(24)

u=xyu = x^{y}.

Is the set on which the function u=u(x;y)u = u(x ; y) is defined:

  1. closed;

  2. open;

  3. path-connected;

  4. a domain;

  5. a closed domain;

  6. convex?

The function u(x;y)u(x ; y) is given by the formula:

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(1)

u=1x2+y2−1u = \frac{1}{x^{2}+y^{2}-1};

(2)

u=xsin⁡yu = \sqrt{x \sin y}

(3)

u=ln⁡(1−2x−x2−y2)+ln⁡(1+2x−x2−y2)u = \ln \left(1-2 x-x^{2}-y^{2}\right)+\ln \left(1+2 x-x^{2}-y^{2}\right);

(4)

u=arcsin⁡(y/x)u = \arcsin (y / x);

(5)

u=xy+arcsin⁡xu = \sqrt{x y}+\arcsin x;

(6)

u=arccos⁡(x/y2)u = \arccos \left(x / y^{2}\right);

(7)

u=arccos⁡(2y+2yx2−1)u = \arccos \left(2 y+2 y x^{2}-1\right).

Find the domain of the function of three variables given by the formula:

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(1)

u=ln⁡(1−x−y−z)u = \ln (1-x-y-z)

(2)

u=1−∣x∣−∣y∣−∣z∣u = \sqrt{1-\left|x\right|-\left|y\right|-\left|z\right|}

(3)

u=1x+1zu = \frac{1}{\sqrt{x}}+\frac{1}{\sqrt{z}}

(4)

u=11−x2−y2−z2u = \frac{1}{\sqrt{1-x^{2}-y^{2}-z^{2}}};

(5)

u=16−x2−z2u = \sqrt{16-x^{2}-z^{2}};

(6)

u=(9−y2−z2)(y2+z2−4)u = \sqrt{\left(9-y^{2}-z^{2}\right)\left(y^{2}+z^{2}-4\right)};

(7)

u=ln⁡(36−36x2−9y2−4z2)u = \ln \left(36-36 x^{2}-9 y^{2}-4 z^{2}\right);

(8)

u=1z−x2−y2u = \frac{1}{\sqrt{z-x^{2}-y^{2}}};

(9)

u=ln⁡(2z2−6x2−3y2−6)u = \ln \left(2 z^{2}-6 x^{2}-3 y^{2}-6\right);

(10)

u=x2+2y2+2yz+z2−2y+1u = \sqrt{x^{2}+2 y^{2}+2 y z+z^{2}-2 y+1};

(11)

u=16−x2−y2−z2ln⁡(x2+y2+z2−4)u = \sqrt{16-x^{2}-y^{2}-z^{2}} \ln \left(x^{2}+y^{2}+z^{2}-4\right);

(12)

u=2(x2+y2+z2)−(x2+y2+z2)2−1u = \sqrt{2\left(x^{2}+y^{2}+z^{2}\right)-\left(x^{2}+y^{2}+z^{2}\right)^{2}-1};

(13)

u=ln⁡x+ln⁡zy−1+ln⁡(5−x−y−z)u = \frac{\ln x+\ln z}{\sqrt{y-1}}+\ln (5-x-y-z);

(14)

u=x+y+z+1−x−y+3x+y−3zu = \sqrt{x}+\sqrt{y}+\sqrt{z}+\sqrt{1-x-y}+\sqrt{3 x+y-3 z};

(15)

u=ln⁡(z2−x2−y2)1−x2−y2−z2u = \frac{\ln \left(z^{2}-x^{2}-y^{2}\right)}{\sqrt{1-x^{2}-y^{2}-z^{2}}};

(16)

u=ln⁡(x2+y2+z2−4)4−x2−y2u = \frac{\ln \left(x^{2}+y^{2}+z^{2}-4\right)}{\sqrt{4-x^{2}-y^{2}}};

(17)

u=ln⁡(4z−z2−x2−y2)4−z−x2−y2u = \frac{\ln \left(4 z-z^{2}-x^{2}-y^{2}\right)}{\sqrt{4-z-x^{2}-y^{2}}};

(18)

u=z−xy+1−z−x2+y2u = \sqrt{z-x y}+\sqrt{1-z-\sqrt{x^{2}+y^{2}}};

(19)

u=arccos⁡(2−x2−y2−z2)u = \arccos \left(2-x^{2}-y^{2}-z^{2}\right);

(20)

u=arcsin⁡x2+y2z−1u = \arcsin \frac{\sqrt{x^{2}+y^{2}}}{z-1};

(21)

u=arccos⁡(x2+y2+z2−3)+arccos⁡x2+y2−3u = \arccos \left(x^{2}+y^{2}+z^{2}-3\right)+\arccos \sqrt{x^{2}+y^{2}-3}.

Prove that the domain of the function given by the formula

u=arcsin⁡xarcsin⁡(2x+y)arcsin⁡(3x+2y+z) u = \arcsin x \arcsin (2 x+y) \arcsin (3 x+2 y+z)

is a closed parallelepiped, and find its vertices.

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Is the set on which the function u(x;y;z)u(x ; y ; z) is defined:

  1. closed;

  2. open;

  3. path-connected;

  4. a domain;

  5. a closed domain;

  6. convex?

The function uu is given by the formula:

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(1)

u=1x2+y2+z2u = \frac{1}{x^{2}+y^{2}+z^{2}};

(2)

u=x+y+zu = \sqrt{x}+\sqrt{y}+\sqrt{z};

(3)

u=ln⁡(xyz)u = \ln (x y z);

(4)

u=z2−x2−y2u = \sqrt{z^{2}-x^{2}-y^{2}};

(5)

y=z2−x2−y2−1y = \sqrt{z^{2}-x^{2}-y^{2}-1};

(6)

u=ln⁡(z2−x2−y2+1)u = \ln \left(z^{2}-x^{2}-y^{2}+1\right);

(7)

u=ln⁡(yln⁡(z−y))u = \ln (y \ln (z-y));

(8)

u=arccos⁡x2+y2z2u = \arccos \frac{x^{2}+y^{2}}{z^{2}};

(9)

u=(xy)zu = (x y)^{z};

(10)

u=zxyu = z^{x y}.

Find the domain and the set of values of the function of four variables given by the formula:

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(1)

u=1/(x12+x22+x32+x42−1)u = 1 /\left(x_{1}^{2}+x_{2}^{2}+x_{3}^{2}+x_{4}^{2}-1\right);

(2)

u=1−x12+4−x22+9−x32+16−x42u = \sqrt{1-x_{1}^{2}}+\sqrt{4-x_{2}^{2}}+\sqrt{9-x_{3}^{2}}+\sqrt{16-x_{4}^{2}};

(3)

u=7−x12−x22−x32−x42−2x3−2x4u = \sqrt{7-x_{1}^{2}-x_{2}^{2}-x_{3}^{2}-x_{4}^{2}-2 x_{3}-2 x_{4}};

(4)

u=ln⁡(144(1−x12)−36x22−16x32−9x42)u = \ln \left(144\left(1-x_{1}^{2}\right)-36 x_{2}^{2}-16 x_{3}^{2}-9 x_{4}^{2}\right).

Find the domain of the function of nn variables given by the formula:

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(1)

u=∑i=1n1−∣xi∣u = \sum_{i = 1}^{n} \sqrt{1-\left|x_{i}\right|}

(2)

u=1−∑i=1nxi+∑i=1nxiu = \sqrt{1-\sum_{i = 1}^{n} x_{i}}+\sum_{i = 1}^{n} \sqrt{x_{i}}

(3)

u=ln⁡(1−∑i=1n(xi−i)2)u = \ln \left(1-\sum_{i = 1}^{n}\left(x_{i}-i\right)^{2}\right);

(4)

u=ln⁡(∑i=1nxi2+∑i<j=1nxixj)u = \ln \left(\sum_{i = 1}^{n} x_{i}^{2}+\sum_{i < j = 1}^{n} x_{i} x_{j}\right)

(5)

u=log⁡a(∑i=1nxi2)u = \sqrt{\log_{a}\left(\sum_{i = 1}^{n} x_{i}^{2}\right)};

(6)

u=∑i=1narcsin⁡(xi−i)u = \sum_{i = 1}^{n} \arcsin \left(x_{i}-i\right).

Find the domain of definition of the function of two variables

u=arccos⁡1x2+y2−1 u = \arccos \frac{1}{x^{2}+y^{2}-1}

and determine whether this function is continuous on its domain of definition.

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References

[Dem97] Demidovich B.P. Сборник задач и упражнений по математическому анализу. 13-е изд., Издательство Московского университета / Издательство "ЧеРо", 1997.
[ККЧ+03] Кудрявцев Л.Д., Кутасов А.Д., Чехлов В.И., Шабунин М.И. Функции многих переменных. 2-е изд., ФИЗМАТЛИТ, 2003.