Double Integration

Double Integration: Solved Examples with Graphs

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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 $$R = [0, 1] \times [0, 2]$$.

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 $$R = [0, 1] \times [0, 2]$$.

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).

Figure 2: Triangle bounded by y = 0, x = 2, and y = x/2

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.

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.

Figure 4: 3D Visualization of the Paraboloid

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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Quick Formulas

Polar $$dA$$: $$r \, dr \, d\theta$$

Volume: $$\iint_D f(x,y) \, dA$$

Average Value: $$\frac{1}{Area(D)} \iint_D f \, dA$$

Mass: $$\iint_D \rho(x,y) \, dA$$