{exm2, cal3} Double Integrals over Rectangle

https://tutorial.math.lamar.edu/Classes/CalcIII/IteratedIntegrals.aspx

Fubini’s Theorem

If $f(x,y)$ is continuous on ($R=[a,b]\times [c,d]$ then, ${\iint }_{R}f(x,y)dA={\int }_a^b{\int }_c^{d}f(x,y)dydx={\int }_c^d({\int }_a^{b}f(x,y)dx)dy$

• These integrals are called iterated integrals
• Where [0,1] is the x-bound, and [0,2] y-bound.
• 2 ways 2 compute (x and y can be either first or last with matching bounds)
• Note how we first compute the inner integral first, then move onto the outer one.

Steps to Solve

1. Compute the inner integral given the requested bounds, treating other var as a constant
2. Compute outer region with requested bounds

Shortening Fact

If $f(x,y)=g(x)h(y)$ and we are integrating over the rectangle $R=[a,b]\times [c,d]$ then,

$${\iint }_{R}f(x,y)dA={\iint }_{R}g(x)h(y)dA=\left({\int }_a^{b}g(x)dx\right)\left({\int }_c^{d}h(y)dy\right)$$

• Estmpsentially, if we have … we can js integrate both separately

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Regions between to curves:

$$f(x,y)=x^2+y$$

We can have:

• Opposite facing parabolas
  • Let $R=$ region between $y=x^2+1,y=1-x-x^2$
  • Then, draw R, and find the max/min/fixed x&y slices and intersections
  • Drawing A we simply take the region between both curves
  • Part B: We setup a SoE:
  $$x^2+1=1-\times -x^{2}2x^2+x=0x(2x+1)=0x=\{0,.5\}(x,y)=(0,1)$$ Then when xy = .5, we get $-\frac{1}{2},\frac{5}{4}$ as our POI (so the x,y point 0,1 and the given point here
  • To find the x max, see its at 0, the min is at $-\frac{1}{2}$
  • y max: $1exm1exm$
• 2 parabolas
• Triangles

Q1, Triangle double \int

Then we want to graph, find bounds + xy min+max Then we integrate. We start given a set of points $(1,-1),(0,0),(0,-1)$

Bounds (called slices in this case)