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Double Integration (rectangular)

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Double Integration (rectangular)
Wize Concept
Iterated Double Integral. If D⊂R2D \subset \mathbb{R}^2D⊂R2 is defined by
g1(x)≤y≤g2(x),a≤x≤b,g_1(x)\leq y\leq g_2(x),\quad a\leq x\leq b,g1​(x)≤y≤g2​(x),a≤x≤b,
where g1(x)g_1(x)g1​(x) and g2(x)g_2(x)g2​(x) are continuous for a≤x≤ba \leq x \leq ba≤x≤b and f(x,y)f(x, y)f(x,y) is continuous on DDD, then
∬Df(x,y)dA=∫ab(∫g1(x)g2(x)f(x,y)dy)dx\iint\limits_{D}f(x,y)dA=\int_a^b\Bigg(\int_{g_1(x)}^{g_2(x)}f(x,y)dy\Bigg)dxD∬​f(x,y)dA=∫ab​(∫g1​(x)g2​(x)​f(x,y)dy)dx
If D⊂R2D \subset \mathbb{R}^2D⊂R2 is defined by
h1(y)≤x≤h2(y),c≤y≤d,h_1(y)\le x\le h_2(y),\quad c\le y\le d,h1​(y)≤x≤h2​(y),c≤y≤d,
where h1(y)h_1(y)h1​(y) and h2(y)h_2(y)h2​(y) are continuous for c≤y≤dc \le y \le dc≤y≤d and f(x,y)f(x, y)f(x,y) is continuous on DDD, then
∬Df(x,y)dA=∫cd(∫h1(y)h2(y)f(x,y)dx)dy\iint\limits_{D}f(x,y)dA=\int_c^d\Bigg(\int_{h_1(y)}^{h_2(y)}f(x,y)dx\Bigg)dyD∬​f(x,y)dA=∫cd​(∫h1​(y)h2​(y)​f(x,y)dx)dy

→\to→ Notes:
It may be useful (or even necessary) to reverse the order of integration. Consider (1) the shape of the region DDD, and (2) the form of the integrand f(x,y)f(x, y)f(x,y).
Strategy
To determine the shape of DDD and/or change the order of integration, eg from dydxdy dxdydx to dxdydx dydxdy:
Using the inequality on the outer variable xxx, sketch the xxx extents of the region.
Sketch the inequalities given by the inner variable yyy within the allowable xxx. This is easiest to do by sketching the boundaries, then determining which side of the line the inequality gives.
Determine the new constant bound on yyy.
Determine the new variable bounds on xxx. Think about a slice with a fixed yyy, and determine the inequality on xxx in terms of yyy.
It is possible to algebraically determine the bounds in the changed order of integration, but a sketch is usually easier.

Practice: Double integration
Compute ∬Dln⁡(ln⁡(x))dA\displaystyle \iint_D\ln(\ln(x))dA∬D​ln(ln(x))dA where DDD is the region bounded by y=0y = 0y=0, xy=1xy = 1xy=1, x=ex=ex=e, and x=e2x=e^2x=e2.

Practice: Changing the Order of a Double Integration
By changing the order of integration, the iterated integral ∫02∫04−x2f(x,y)dy dx\displaystyle\int_0^2\int_0^{4-x^2} 
  f (x, y)dy\,dx∫02​∫04−x2​f(x,y)dydx is the same  as:   

∫04∫04−y2f(x,y)dxdy\int_0^4\int_0^{4-y^2}f\left(x,y\right)dxdy∫04​∫04−y2​f(x,y)dxdy

∫02∫04−y2f(x,y)dxdy\int_0^2\int_0^{4-y^2}f\left(x,y\right)dxdy∫02​∫04−y2​f(x,y)dxdy

∫04∫04−yf(x,y)dxdy\int_0^4\int_0^{\sqrt{4-y}}f\left(x,y\right)dxdy∫04​∫04−y​​f(x,y)dxdy

∫02∫04−yf(x,y)dxdy\int_0^2\int_0^{\sqrt{4-y}}f\left(x,y\right)dxdy∫02​∫04−y​​f(x,y)dxdy

None of the above

I don't know

Practice: Double Integral
Evaluate
∫01∫x1ey2dydx\int_0^1\int_x^1e^{y^2}dydx∫01​∫x1​ey2dydx

Interpretation of a Double Integral
Wize Concept
Interpretation of Double Integral. We can interpret the double integral as:
1. Area. If we fix f(x,y)=1f(x, y) = 1f(x,y)=1 for all (x,y)∈D(x, y) \in D(x,y)∈D, then the double integral can be interpreted as the area of the set DDD:
A(D)=∬D1dAA(D)=\iint\limits_{D}1dAA(D)=D∬​1dA
2. Volume. If f(x,y)≥0f(x, y)\geq0f(x,y)≥0 for all (x,y)∈D(x, y) \in D(x,y)∈D, then the double integral can be interpreted as the volume of the 3−D3-D3−D set defined by
S={(x,y,z):0≤z≤f(x,y), (x,y)∈D},S=\{(x,y,z):0\le z\le f(x,y),\ (x,y)\in D\},S={(x,y,z):0≤z≤f(x,y), (x,y)∈D},
which is the solid region of height f(x,y)f(x, y)f(x,y) and base DDD.
V(S)=∬Df(x,y)dA.V(S)=\iint\limits_{D}f(x,y)dA.V(S)=D∬​f(x,y)dA.

Practice: Volume under a surface
Find the volume under the surface z=−xx2+y23z=-\frac{x}{\sqrt[3]{x^2+y^2}}z=−3x2+y2​x​ and above the triangle formed by y=xy = xy=x, y=−2y = -2y=−2, and the yyy-axis.

97(2−33)\frac{9}{7}\left(2-\sqrt[3]{3}\right)79​(2−33​)

9(2−33)9\left(2-\sqrt[3]{3}\right)9(2−33​)

97(43)\frac{9}{7}\left(\sqrt[3]{4}\right)79​(34​)

97(2−43)\frac{9}{7}\left(2-\sqrt[3]{4}\right)79​(2−34​)

None of the above

I don't know

Extra Practice

Practice: Double Integrals over General Regions (2)

Evaluate ∫D∫(6x2−40y)dA \displaystyle\int_D\int\Bigg(6x^2-40y\Bigg)dA~∫D​∫(6x2−40y)dA where D is the triangle with vertices (0, 3), (1, 1), and (5, 3). 

Practice: Changing the order of integration

Evaluate
∫01∫x1ey2dydx\int_0^1\int_x^1e^{y^2}dydx∫01​∫x1​ey2dydx

Practice: Double Integrals over General Regions (2)

Evaluate ∫D∫(6x2−40y)dA \displaystyle\int_D\int\Bigg(6x^2-40y\Bigg)dA~∫D​∫(6x2−40y)dA where D is the triangle with vertices (0, 3), (1, 1), and (5, 3). 

Practice: Double integration

Compute ∬Dln⁡(ln⁡(x))dA\displaystyle \iint_D\ln(\ln(x))dA∬D​ln(ln(x))dA where DDD is the region bounded by y=0y = 0y=0, xy=1xy = 1xy=1, x=ex=ex=e, and x=e2x=e^2x=e2.