?Mean Value Theorem for Double Integrals The Mean Value Theorem for double integrals

Evaluate the iterated integral.

Evaluate the iterated integral. \(\int_{0}^{2} \int_{0}^{y^{2}} x^{2} y \ d x \ d y\) Equation Transcription: Text Transcription: integral _0 ^2 integral _0 ^y^2 x^2 y dx dy

Evaluate the iterated integral.

Evaluate the iterated integral. \(\int_{0}^{\pi / 2 x} \int_{0}^{x} x \sin y d y d x\) Equation Transcription: Text Transcription: int_0^pi/2 \int_0^{x} x sin y dy dx

Evaluate the iterated integral. \(\int_{0}^{1} \int_{0}^{2} \cos \left(s^{3}\right) d t d s\) Equation Transcription: Text Transcription: int_0^1 int_0^2 cos (s^3) dt ds

Evaluate the iterated integral.

(a) Express the double integral as an iterated integral for the given function and region . (b) Evaluate the iterated integral.

(a) Express the double integral as an iterated integral for the given function and region . (b) Evaluate the iterated integral.

(a) Express the double integral as an iterated integral for the given function and region . (b) Evaluate the iterated integral.

(a) Express the double integral \(\iint_{D} f(x, y) d A\) as an iterated integral for the given function f and region D. (b) Evaluate the iterated integral. \(f(x, y)=x\) Equation Transcription: ? Text Transcription: double integral_D f(x, y) dA f(x, y)=x

Evaluate the double integral. \(\iint_{D} \frac{y}{x^{2}+1} d A, D=\{(x, y) \mid 0 \leqslant x \leqslant 4,0 \leqslant y \leqslant \sqrt{x}\}\) Equation Transcription: ???? Text Transcription: double integral y/x^2 +1 dA, D={(x, y)|0 leqslant x leqslant 4, 0 leqslant y leqslant sqrt x}

Evaluate the double integral.

Evaluate the double integral. \(\iint_{D} e^{-y^{2}} d A, D=\{(x, y) \mid 0 \leqslant y \leqslant 3,0 \leqslant x \leqslant y\}\) Equation Transcription: ???? Text Transcription: double integral_D e^-y^2 dA, D={(x,y)|0 leqslant y leqslant 3, 0 leqslant x leqslant y }

Evaluate the double integral. \(\iint_{D} y \sqrt{x^{2}-y^{2}} d A, D=\{(x, y) \mid 0 \leqslant x \leqslant 2,0 \leqslant y \leqslant x\}\) Equation Transcription: ???? Text Transcription: double integral_D y sqrt x^2 -y^2 dA, D={(x, y)|0 leqslant x leqslant 2, 0 leqslant y leqslant x }

Draw an example of a region that is (a) type I but not type II (b) type II but not type I

Draw an example of a region that is (a) both type I and type II (b) neither type I nor type II

Express D as a region of type I and also as a region of type II. Then evaluate the double integral in two ways. is enclosed by the lines

Express D as a region of type I and also as a region of type II. Then evaluate the double integral in two ways. is enclosed by the curves

Set up iterated integrals for both orders of integration. Then evaluate the double integral using the easier order and explain why it’s easier. \(\iint_{D} y d A, D\) is bounded by \(y=x-2, x=y^{2}\) Equation Transcription: Text Transcription: double integral_D ydA, D y =x-2, x=y^2

Set up iterated integrals for both orders of integration. Then evaluate the double integral using the easier order and explain why it’s easier. \(\iint_{D} y^{2} e^{x y} d A, D\) is bounded by \(y=x, y=4, x=0\) Equation Transcription: Text Transcription: double integral_D y^2 e^xy dA, D y = x, y = 4, x = 0

Set up iterated integrals for both orders of integration. Then evaluate the double integral using the easier order and explain why it's easier. \(\iint_D \sin ^2 x d A\), D is bounded by y=cos x, \(0 \leqslant x \leqslant \pi / 2\), y=0, x=0

Set up iterated integrals for both orders of integration. Then evaluate the double integral using the easier order and explain why it’s easier. \(\iint_{D} 6 x^{2} d A, D\) is bounded by \(y=x^{3}, y=2 x+4, x=0\) Equation Transcription: Text Transcription: double integral_D 6x^2 dA, D y=x^3 ,y=2x +4, x =0

Evaluate the double integral. \(\iint_{D} x \cos y d A\), \(D\) is bounded by \(y=0, y=x^{2}, x=1\) Equation Transcription: Text Transcription: double integral_D x cos y dA D y = 0, y = x^2, x = 1

Evaluate the double integral. is bounded by

Evaluate the double integral. \(\iint_{D} y^{2} d A\) D is the triangular region with vertices (0, 1), (1, 2), (4, 1)

Evaluate the double integral. \(\iint_{D} x y d A, D\) is enclosed by the quarter-circle \(y=\sqrt{1-x^{2}}, x \geqslant 0\), and the axes Equation Transcription: ? Text Transcription: double integral_D xy dA, D y = sqrt 1-x^2 , x geqslant 0

Evaluate the double integral. , is bounded by the circle with center the origin and radius 2

Evaluate the double integral. \(\iint_{D} y d A,\) D is the triangular region with vertices (0, 0), (1, 1), and (4, 0)

The figure shows a surface and a region D in the xy-plane. (a) Set up an iterated double integral for the volume of the solid that lies under the surface and above D. (b) Evaluate the iterated integral to find the volume of the solid.

The figure shows a surface and a region D in the xy-plane. (a) Set up an iterated double integral for the volume of the solid that lies under the surface and above D. (b) Evaluate the iterated integral to find the volume of the solid.

Find the volume of the given solid. Under the plane and above the region enclosed by the parabolas and

Find the volume of the given solid. Under the surface and above the region enclosed by and

Find the volume of the given solid. Under the surface and above the triangle with vertices , and

Find the volume of the given solid. Enclosed by the paraboloid and the planes , and

Find the volume of the given solid. The tetrahedron enclosed by the coordinate planes and the plane

Find the volume of the given solid. Bounded by the planes , and

Find the volume of the given solid. Enclosed by the cylinders and the planes

Find the volume of the given solid. Bounded by the cylinder and the planes , in the first octant

Find the volume of the given solid. Bounded by the cylinder and the planes , in the first octant

Find the volume of the given solid. Bounded by the cylinders and

Use a graph to estimate the -coordinates of the points of intersection of the curves and If is the region bounded by these curves, estimate

Find the approximate volume of the solid in the first octant that is bounded by the planes \(y=x, z=0\), and \(z=x\) and the cylinder \(y=\cos x\). (Use a graph to estimate the points of intersection.) Equation Transcription: Text Transcription: y = x, z = 0 z = x y = cos x

Find the volume of the solid by subtracting two volumes. The solid enclosed by the parabolic cylinders , and the planes

Find the volume of the solid by subtracting two volumes. The solid enclosed by the parabolic cylinder \(y=x^{2}\) and the planes \(z=3 y, \ z=2+y\)

Find the volume of the solid by subtracting two volumes. The solid under the plane , above the plane , and between the parabolic cylinders and

Find the volume of the solid by subtracting two volumes. The solid in the first octant under the plane \(z=x+y\), above the surface \(z=x y\), and enclosed by the surfaces \(x=0, \ y=0\), and \(x^{2}+y^{2}=4\).

Sketch the solid whose volume is given by the iterated integral.

Sketch the solid whose volume is given by the iterated integral. \(\int_{0}^{1} \int_{0}^{1-x^{2}}(1-x) d y d x\) Equation Transcription: Text Transcription: integral _0 ^1 integral _0 ^1-x^2 (1-x) dy dx

Sketch the solid whose volume is given by the iterated integral. \(\int_{0}^{3} \int_{0}^{y} \sqrt{9-x^{2}} d x d y\) Equation Transcription: Text Transcription: integral _0 ^3 integral _0 ^y sqrt 9-x^2 dx dy

Sketch the solid whose volume is given by the iterated integral. \(\int_{-2}^{2} \int_{-1}^{3-x^{2}} e^{-y} \ d y \ d x\)

Use a computer algebra system to find the exact volume of the solid. Under the surface and above the region bounded by the curves and for

Use a computer algebra system to find the exact volume of the solid. Between the paraboloids and and inside the cylinder

Use a computer algebra system to find the exact volume of the solid. Enclosed by and

Use a computer algebra system to find the exact volume of the solid. Enclosed by and

Sketch the region of integration and change the order of integration.

Sketch the region of integration and change the order of integration. \(\int_{0}^{2} \int_{x^{2}}^{4} f(x, y) d y d x\) Equation Transcription: Text Transcription: integral _0 ^2 integral _x^2 ^4 f(x, y) dy dx

Sketch the region of integration and change the order of integration. \(\int_{0}^{\pi / 2} \int_{\sin x}^{1} f(x, y) d y d x\) Equation Transcription: Text Transcription: \int_0^\pi/2 \int_sin x^1 f(x,y) dy dx

Sketch the region of integration and change the order of integration. \(\int_{-2}^{2} \int_{0}^{\sqrt{4-y^{2}}} f(x, y) d x d y\) Equation Transcription: Text Transcription: integral _-2 ^2 integral _0 ^sqrt 4-y^2 f(x, y) dx dy

Sketch the region of integration and change the order of integration. \(\int_{1}^{2} \int_{0}^{\ln x} f(x, y) d y d x\) Equation Transcription: Text Transcription: int_1^2 \int_0^ln x f(x,y) dy dx

Sketch the region of integration and change the order of integration. \(\int_{0}^{1} \int_{\text {arctan } x}^{\pi / 4} f(x, y) d y d x\)

Evaluate the integral by reversing the order of integration. \(\int_{0}^{1} \int_{3 y}^{3} e^{x^{2}} d x d y\) Equation Transcription: Text Transcription: integral _0 ^1 integral _3y ^3 e^x^2 dx dy

Evaluate the integral by reversing the order of integration. sin y dy dx

Evaluate the integral by reversing the order of integration. \(\int_{0}^{1} \int_{\sqrt{x}}^{1} \sqrt{y^{3}+1} \ d y \ d x\)

Evaluate the integral by reversing the order of integration. \(\int_{0}^{2} \int_{y / 2}^{1} y \cos \left(x^{3}-1\right) \ d x \ d y\)

Evaluate the integral by reversing the order of integration. cos x dx dy

Evaluate the integral by reversing the order of integration. dx dy

Express D as a union of regions of type I or type II and evaluate the integral.

Express D as a union of regions of type I or type II and evaluate the integral.

Use Property 10 to estimate the value of the integral.

Use Property 10 to estimate the value of the integral.

Find the average value of f over the region D. f(x, y) = xy, D is the triangle with vertices (0, 0), (1, 0), and (1, 3)

Find the average value of f over the region D. f(x, y) = x sin y, D is enclosed by the curves y = 0, y = x2, and x = 1

Prove Property 10.

In evaluating a double integral over a region D, a sum of iterated integrals was obtained as follows: Sketch the region D and express the double integral as an iterated integral with reversed order of integration.

Use geometry or symmetry, or both, to evaluate the double integral.

Use geometry or symmetry, or both, to evaluate the double integral. D is the disk with center the origin and radius R

Use geometry or symmetry, or both, to evaluate the double integral. D is the rectangle 0 ? x ? a, 0 ? y ? b

Use geometry or symmetry, or both, to evaluate the double integral.

Use geometry or symmetry, or both, to evaluate the double integral. \(\begin{aligned}&\iint_D\left(a x^3+b y^3+\sqrt{a^2-x^2}\right) d A, \\&D=[-a, a]\times[-b, b]\end{aligned}\)

Mean Value Theorem for Double Integrals The Mean Value Theorem for double integrals says that if \(f\) is a continuous function on a plane region \(D\) that is of type I or type II, then there exists a point \(\left(x_{0}, y_{0}\right)\) in \(D\) such that \(\iint_{D} f(x, y) d A=f\left(x_{0}, y_{0}\right) \ A(D)\) Use the Extreme Value Theorem (14.7.8) and Property 15.2.10 of integrals to prove the Mean Value Theorem for double integrals. (Use the proof of the single-variable version in Section 6.5 as a guide.)

Mean Value Theorem for Double Integrals The Mean Value Theorem for double integrals says that if f is a continuous function on a plane region D that is of type I or type II, then there exists a point (x0, y0) in D such that Suppose that f is continuous on a disk that contains the point (a, b). Let Dr be the closed disk with center (a, b) and radius r. Use the Mean Value Theorem for double integrals to show that

Graph the solid bounded by the plane \(x+y+z=1\) and the paraboloid \(z=4-x^{2}-y^{2}\) and find its exact volume. (Use a computer algebra system to find the equations of the boundary curves of the region of integration and to evaluate the double integral.)