Understanding Double Integration
Double integration extends the concept of a single integral to functions of two variables, $$f(x, y)$$. Geometrically, it represents the volume under a surface within a given region $$D$$.
Example 1: Rectangular Bounds
Evaluate $$\iint_R (x + 2y) \, dA$$ over $$R = [0, 1] \times [0, 2]$$.
x
y
Step 1: Setup Iterated Integral
$$\int_{0}^{1} \int_{0}^{2} (x + 2y) \, dy \, dx$$
Step 2: Integrate w.r.t $y$
$$\int_{0}^{1} [xy + y^2]_{0}^{2} \, dx = \int_{0}^{1} (2x + 4) \, dx$$
Step 3: Integrate w.r.t $x$
$$[x^2 + 4x]_{0}^{1} = 1 + 4 = 5$$
Final Answer: 5
Example 2: Polar Transformation
Find the area of a circle with radius $$a$$.
$$\iint_D dA = \int_{0}^{2\pi} \int_{0}^{a} r \, dr \, d\theta$$
Step 1: Radial Integral
$$\int_{0}^{a} r \, dr = \left[ \frac{r^2}{2} \right]_{0}^{a} = \frac{a^2}{2}$$
Step 2: Angular Integral
$$\int_{0}^{2\pi} \frac{a^2}{2} \, d\theta = \left[ \frac{a^2 \theta}{2} \right]_{0}^{2\pi} = \pi a^2$$
Final Answer: $$\pi a^2$$
Example 3: Triangular Region
Evaluate
\[
\iint_R xy\,dA
\]
where \(R\) is bounded by \(y=0,\; y=x,\; x=2\).
\[
\int_0^2\int_0^x xy\,dy\,dx
= \int_0^2 \frac{x^3}{2}\,dx
= 2
\]
Answer: \( \boxed{2} \)
Example 4: Area Between Curves
Find the area bounded by \(y=x^2\) and \(y=4\).
\[
\text{Area} = \int_{-2}^{2}\int_{x^2}^{4} dy\,dx
= \int_{-2}^{2} (4-x^2)\,dx
= \frac{32}{3}
\]
Answer: \( \boxed{\frac{32}{3}} \)
Example 5: Circular Region
Evaluate
\[
\iint_R (x^2+y^2)\,dA
\]
where \(x^2+y^2 \le 1\).
Using polar coordinates:
\[
\int_0^{2\pi}\int_0^1 r^3\,dr\,d\theta
= \frac{\pi}{2}
\]
Answer: \( \boxed{\frac{\pi}{2}} \)
Example 6: Region Between Two Curves
Evaluate
\[
\iint_R y\,dA
\]
where \(y=x^2\) and \(y=x\).
\[
\int_0^1\int_{x^2}^{x} y\,dy\,dx
= \int_0^1 \frac{x^2-x^4}{2}\,dx
= \frac{1}{15}
\]
Answer: \( \boxed{\frac{1}{15}} \)
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Double Integration: 7 Solved Examples with Step-by-Step Solutions
Double integration is used to calculate the area of a 2D region or the volume under a 3D surface. Here are four essential examples.
Example 1: Rectangular Region
Evaluate $$\iint_R (8x + 6y) \, dA$$ where $$R = [0, 1] \times [0, 2]$$.
y=2
x=1
Figure 1: Rectangular Domain on the xy-plane
Step 1: Setup
$$\int_{0}^{2} \int_{0}^{1} (8x + 6y) \, dx \, dy$$
Step 2: Inner Integration (dx)
$$[4x^2 + 6xy]_{0}^{1} = 4 + 6y$$
Step 3: Outer Integration (dy)
$$\int_{0}^{2} (4 + 6y) \, dy = [4y + 3y^2]_{0}^{2} = 8 + 12 = 20$$
Result: 20
Example 8: Triangular Region (Type I)
Evaluate $$\iint_D xy^2 \, dA$$ for the triangle with vertices (0,0), (2,0), (2,1).
Step 1: Setup
$$\int_{0}^{2} \int_{0}^{x/2} xy^2 \, dy \, dx$$
Step 2: Inner Integration (dy)
$$[x \frac{y^3}{3}]_{0}^{x/2} = \frac{x}{3} \cdot \frac{x^3}{8} = \frac{x^4}{24}$$
Step 3: Outer Integration (dx)
$$\int_{0}^{2} \frac{x^4}{24} \, dx = [\frac{x^5}{120}]_{0}^{2} = \frac{32}{120} = 4/15$$
Result: 4/15
Example 9: Polar Coordinates
Evaluate $$\iint_D (x^2 + y^2) \, dA$$ where $$D$$ is the unit circle.
r=1
Figure 3: Unit Circle in Polar Form ($r=1, \theta=2\pi$)
Step 1: Setup (Polar)
$$\int_{0}^{2\pi} \int_{0}^{1} (r^2)r \, dr \, d\theta$$
Step 2: Solve
$$\int_{0}^{2\pi} [\frac{r^4}{4}]_0^1 \, d\theta = \int_{0}^{2\pi} \frac{1}{4} d\theta = \frac{2\pi}{4} = \pi/2$$
Result: π/2
Example 10: Volume of Paraboloid
Find volume under $$z = 1 - x^2 - y^2$$ above the xy-plane.
Solution:
Using polar conversion $$1-(x^2+y^2) = 1-r^2$$:
$$\int_{0}^{2\pi} \int_{0}^{1} (1-r^2)r \, dr \, d\theta = \pi/2$$
Result: π/2
Graphs generated for educational purposes. For interactive 3D plots, use GeoGebra 3D .
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