Double integrals, change of order of integration, change of variables

Explanation:

We are tasked with evaluating the double integral:

\[ \int_0^1 \int_0^2 (x + y) \, dx \, dy \]

Step 1: First, integrate with respect to \(x\):

\[ \int_0^2 (x + y) \, dx = \left[ \frac{x^2}{2} + yx \right]_0^2 = (2 + 2y) - 0 = 2 + 2y \]

Step 2: Now, integrate with respect to \(y\):

\[ \int_0^1 (2 + 2y) \, dy = \left[ 2y + y^2 \right]_0^1 = (2 + 1) - 0 = 3 \]

Therefore, the value of the integral is 3.

Explanation:

First find intersection points by solving x² = 2x - x²:

\[ 2x² - 2x = 0 \Rightarrow x = 0, 1 \]

The area is calculated as:

\[ \int_0^1 \int_{x²}^{2x-x²} dy\,dx = \int_0^1 (2x - 2x²)\,dx = \left[x² - \frac{2x³}{3}\right]_0^1 = 1 - \frac{2}{3} = \frac{1}{3} \]

Explanation:

Convert to polar coordinates where x² + y² = r²:

\[ \int_0^{2π} \int_0^2 (4 - r²) r\,dr\,dθ \]

First integrate with respect to r:

\[ \int_0^2 (4r - r³)\,dr = \left[2r² - \frac{r⁴}{4}\right]_0^2 = 8 - 4 = 4 \]

Then integrate with respect to θ:

\[ \int_0^{2π} 4\,dθ = 8π \]

Explanation:

The volume is given by the triple integral:

\[ \int_0^1 \int_0^1 \int_0^1 dz\,dy\,dx = 1 \times 1 \times 1 = 1 \]

Each integral evaluates to 1, and their product is 1.

Explanation:

In cylindrical coordinates:

\[ \int_0^{2π} \int_0^2 \int_r^2 r\,dz\,dr\,dθ \]

First integrate with respect to z:

\[ \int_r^2 dz = 2 - r \]

Then integrate with respect to r:

\[ \int_0^2 r(2 - r)\,dr = \left[r² - \frac{r³}{3}\right]_0^2 = 4 - \frac{8}{3} = \frac{4}{3} \]

Finally integrate with respect to θ:

\[ \int_0^{2π} \frac{4}{3}\,dθ = \frac{8π}{3} \]

Explanation:

To find the area between the curves:

\[ \text{Area} = \int_0^1 \int_{x^2}^x dy\,dx \]

First integrate with respect to y:

\[ \int_{x^2}^x dy = x - x^2 \]

Then integrate with respect to x:

\[ \int_0^1 (x - x^2)\,dx = \left[\frac{x^2}{2} - \frac{x^3}{3}\right]_0^1 = \frac{1}{2} - \frac{1}{3} = \frac{1}{6} \]

Explanation:

Using polar coordinates:

\[ \text{Volume} = \int_0^{2π} \int_0^1 (1 - r^2) r\,dr\,dθ \]

First integrate with respect to r:

\[ \int_0^1 (r - r^3)\,dr = \left[\frac{r^2}{2} - \frac{r^4}{4}\right]_0^1 = \frac{1}{2} - \frac{1}{4} = \frac{1}{4} \]

Then integrate with respect to θ:

\[ \int_0^{2π} \frac{1}{4}\,dθ = \frac{π}{2} \]

Explanation:

The volume is given by:

\[ \int_0^1 \int_0^{1-x} \int_0^{1-x-y} dz\,dy\,dx \]

First integrate with respect to z:

\[ \int_0^{1-x-y} dz = 1 - x - y \]

Then integrate with respect to y:

\[ \int_0^{1-x} (1 - x - y)\,dy = \left[(1-x)y - \frac{y^2}{2}\right]_0^{1-x} = \frac{(1-x)^2}{2} \]

Finally integrate with respect to x:

\[ \int_0^1 \frac{(1-x)^2}{2}\,dx = \left[-\frac{(1-x)^3}{6}\right]_0^1 = \frac{1}{6} \]

Explanation:

First find intersection points:

\[ 2cosθ = 1 \Rightarrow θ = ±π/3 \]

The area is:

\[ \frac{1}{2}\int_{-π/3}^{π/3} [(2cosθ)^2 - 1^2]\,dθ = \frac{1}{2}\int_{-π/3}^{π/3} (4cos²θ - 1)\,dθ \]

Using the identity cos²θ = (1 + cos2θ)/2:

\[ = \frac{1}{2}\int_{-π/3}^{π/3} (1 + 2cos2θ)\,dθ = \frac{1}{2}\left[θ + sin2θ\right]_{-π/3}^{π/3} = π/3 + √3/2 \]

Explanation:

In cylindrical coordinates (x = rcosθ, y = rsinθ):

\[ \text{Volume} = \int_0^π \int_0^{2sinθ} \int_{r^2}^{2rsinθ} r\,dz\,dr\,dθ \]

First integrate with respect to z:

\[ \int_{r^2}^{2rsinθ} dz = 2rsinθ - r^2 \]

Then integrate with respect to r:

\[ \int_0^{2sinθ} r(2rsinθ - r^2)\,dr = \left[\frac{2r^3sinθ}{3} - \frac{r^4}{4}\right]_0^{2sinθ} = \frac{4sin^4θ}{3} \]

Finally integrate with respect to θ:

\[ \int_0^π \frac{4sin^4θ}{3}\,dθ = \frac{4}{3} \cdot \frac{3π}{8} = \frac{π}{2} \]