Use a transformation to evaluate the double integral of f(x,y) given by f(x,y)=cos(2x-y)sin(x+2y) over the square region with vertices at
(0,0) (1,-2) (3,-1) & (2,1) (My notes from class-uses substitution, change of variables)

I have let u=(2x-y) & v=(x+2y) using substitution (change of variables)

Solution Summary

A square is defined as a matrix of derivatives in this elegant solution to a double integral.

I have a doubleintegral of the form
I from 0 to 2 and I from 0 to y (4-y^2) dx dy
How does one convert this doubleintegral into a triple integral?
keywords: integration, integrates, integrals, integrating, double, triple, multiple

Consider the vector field F=((x^2)*y+(y^3)/3)i,(i is the horizontal unit vector) and let C be the portion of the graph y=f(x) running from (x1,f(x1)) to (x2,f(x2)) (assume that x1integral "integral(F.dr)" is equal to the polar moment of inertia of the region R lying below

Please see the attached file for the fully formatted problems.
1. Use an iterated integral to find the area of a region...
2. Evaluate the doubleintegral...
3. Use doubleintegral to find the volume of a solid...
4. Verify moments of inertia...
5. Limit of doubleintegral...
6. Surface area...
7. Triple integral...

Find the volume of the following region in space: The first octant region bounded by the coordinate planes and the surfaces y=1-x^2, z=1-x^2.
This question is #12 (section 9.3) in Advanced engineering mathmatics (8th ed.) by Kreyszig.
This section deals with the evaluation of doubleintegrals.

See the attached file for full description.
26. Evaluate the triple integral, where E is bounded by the planes y = 0, z = 0, x + y = 2 and the cylinder y^2 + z^2 =1 in the first octant.
Find the volume of the given solid
30. Under the surface z = x^2y and above triangle in the xy-plane with vertices (1, 0), (2, 1), and (4

Please help with the following problem. Provide step by step calculations for each.
The average value of f(x) = 1/x on the interval [4, 16] is
(ln 2)/3
(ln 2)/6
(ln 2)/12
3/2
0
1
none of these
Find the area, in square units, of the region b

Use double integration in polar coordinates to find the volume of the solid that lies below the given surface and above the plane region R bounded by the given curve.
1. z=x^2+y^2; r=3
Evaluate the given integral by first converting to polar coordinates.
2. ∬_(0,x)^1,1▒〖x^2 dy dx〗
Solve by double i