1. **Problem statement:** Find the area of region T bounded by the y-axis. 2. **Given:** Region T is bounded by the y-axis and some curve (not explicitly given here, but assumed from context). 3. **Formula for area bounded by y-axis:** If the region is bounded by the y-axis (x=0) and a curve y=f(x), the area can be found by integrating with respect to x from 0 to the curve's boundary. 4. **Step:** Identify the limits of integration and the function defining the boundary of region T. 5. **Step:** Calculate the definite integral $$\text{Area}_T = \int_0^a f(x) \, dx$$ where $a$ is the x-boundary of region T. 6. **Note:** Since the exact function and limits are not provided in the question snippet, the exact integral cannot be computed here. --- 1. **Problem statement:** Find the area of region U. 2. **Step:** Identify the function and limits defining region U. 3. **Step:** Calculate the definite integral for the area. --- 1. **Problem statement:** Find the area of region V. 2. **Step:** Identify the function and limits defining region V. 3. **Step:** Calculate the definite integral for the area. --- 1. **Problem statement:** Find the area of region W bounded by the x-axis and y-axis. 2. **Step:** The area bounded by the x-axis (y=0) and y-axis (x=0) and a curve can be found by integrating the function over the appropriate limits. 3. **Step:** Calculate $$\text{Area}_W = \int_0^b f(x) \, dx$$ where $b$ is the x-boundary. --- 1. **Problem statement:** Let R be the region in the first and second quadrants bounded above by $$y=\frac{20}{1+x^2}$$ and below by $$y=2$$. 2. **Find the area of R:** - The area between two curves $$y=f(x)$$ and $$y=g(x)$$ from $$x=a$$ to $$x=b$$ is $$\int_a^b (f(x)-g(x)) \, dx$$. - Here, $$f(x) = \frac{20}{1+x^2}$$ and $$g(x) = 2$$. - Find the intersection points by solving $$\frac{20}{1+x^2} = 2$$: $$20 = 2(1+x^2) \Rightarrow 20 = 2 + 2x^2 \Rightarrow 18 = 2x^2 \Rightarrow x^2 = 9 \Rightarrow x = \pm 3$$. - So, the limits are $$-3$$ to $$3$$. - Area $$= \int_{-3}^3 \left( \frac{20}{1+x^2} - 2 \right) dx$$. 3. **Calculate the integral:** $$\int_{-3}^3 \frac{20}{1+x^2} dx - \int_{-3}^3 2 \, dx = 20 \int_{-3}^3 \frac{1}{1+x^2} dx - 2 \times (3 - (-3))$$ - Since $$\frac{1}{1+x^2}$$ is even, $$\int_{-3}^3 \frac{1}{1+x^2} dx = 2 \int_0^3 \frac{1}{1+x^2} dx$$. - So, $$= 20 \times 2 \int_0^3 \frac{1}{1+x^2} dx - 2 \times 6 = 40 \int_0^3 \frac{1}{1+x^2} dx - 12$$ - The integral $$\int \frac{1}{1+x^2} dx = \arctan x$$. - Evaluate: $$40 [\arctan(3) - \arctan(0)] - 12 = 40 \arctan(3) - 12$$ 4. **Final area:** $$\boxed{40 \arctan(3) - 12}$$ --- 1. **Find the volume generated when R is rotated about the x-axis:** - Use the disk method: volume $$= \pi \int_a^b \left( f(x)^2 - g(x)^2 \right) dx$$ where $$f(x)$$ is the outer radius and $$g(x)$$ is the inner radius. - Here, outer radius $$= \frac{20}{1+x^2}$$, inner radius $$= 2$$, limits $$-3$$ to $$3$$. - Volume $$= \pi \int_{-3}^3 \left( \left( \frac{20}{1+x^2} \right)^2 - 2^2 \right) dx = \pi \int_{-3}^3 \left( \frac{400}{(1+x^2)^2} - 4 \right) dx$$. - Split integral: $$= \pi \left( \int_{-3}^3 \frac{400}{(1+x^2)^2} dx - \int_{-3}^3 4 \, dx \right) = \pi \left( 400 \int_{-3}^3 \frac{1}{(1+x^2)^2} dx - 4 \times 6 \right)$$ - Since $$\frac{1}{(1+x^2)^2}$$ is even, $$= \pi \left( 400 \times 2 \int_0^3 \frac{1}{(1+x^2)^2} dx - 24 \right) = \pi (800 I - 24)$$ where $$I = \int_0^3 \frac{1}{(1+x^2)^2} dx$$. - Use formula: $$\int \frac{dx}{(1+x^2)^2} = \frac{x}{2(1+x^2)} + \frac{\arctan x}{2} + C$$. - Evaluate from 0 to 3: $$I = \left[ \frac{x}{2(1+x^2)} + \frac{\arctan x}{2} \right]_0^3 = \left( \frac{3}{2(1+9)} + \frac{\arctan 3}{2} \right) - 0 = \frac{3}{20} + \frac{\arctan 3}{2}$$ - Substitute back: $$\text{Volume} = \pi \left( 800 \left( \frac{3}{20} + \frac{\arctan 3}{2} \right) - 24 \right) = \pi \left( 800 \times \frac{3}{20} + 800 \times \frac{\arctan 3}{2} - 24 \right)$$ - Simplify: $$= \pi \left( 120 + 400 \arctan 3 - 24 \right) = \pi (96 + 400 \arctan 3)$$ 5. **Final volume:** $$\boxed{\pi (96 + 400 \arctan 3)}$$ --- 1. **Find the volume of the solid with base R and semicircular cross sections perpendicular to x-axis:** - Cross-sectional area $$A(x) = \frac{\pi}{8} [f(x) - g(x)]^2$$ where $$f(x)$$ and $$g(x)$$ are the upper and lower boundaries. - Here, $$f(x) = \frac{20}{1+x^2}$$, $$g(x) = 2$$. - Volume $$= \int_{-3}^3 A(x) dx = \int_{-3}^3 \frac{\pi}{8} \left( \frac{20}{1+x^2} - 2 \right)^2 dx$$. - Simplify: $$= \frac{\pi}{8} \int_{-3}^3 \left( \frac{20}{1+x^2} - 2 \right)^2 dx$$. - Since the integrand is even, rewrite: $$= \frac{\pi}{8} \times 2 \int_0^3 \left( \frac{20}{1+x^2} - 2 \right)^2 dx = \frac{\pi}{4} \int_0^3 \left( \frac{20}{1+x^2} - 2 \right)^2 dx$$. - Expand the square: $$\left( \frac{20}{1+x^2} - 2 \right)^2 = \left( \frac{20}{1+x^2} \right)^2 - 2 \times 2 \times \frac{20}{1+x^2} + 2^2 = \frac{400}{(1+x^2)^2} - \frac{80}{1+x^2} + 4$$. - Integral becomes: $$\int_0^3 \left( \frac{400}{(1+x^2)^2} - \frac{80}{1+x^2} + 4 \right) dx = 400 \int_0^3 \frac{1}{(1+x^2)^2} dx - 80 \int_0^3 \frac{1}{1+x^2} dx + 4 \times 3$$ - Use previous results: $$\int_0^3 \frac{1}{(1+x^2)^2} dx = \frac{3}{20} + \frac{\arctan 3}{2}$$ $$\int_0^3 \frac{1}{1+x^2} dx = \arctan 3$$ - Substitute: $$= 400 \left( \frac{3}{20} + \frac{\arctan 3}{2} \right) - 80 \arctan 3 + 12 = 60 + 200 \arctan 3 - 80 \arctan 3 + 12 = 72 + 120 \arctan 3$$ - Multiply by $$\frac{\pi}{4}$$: $$\text{Volume} = \frac{\pi}{4} (72 + 120 \arctan 3) = 18 \pi + 30 \pi \arctan 3$$ 6. **Final volume:** $$\boxed{18 \pi + 30 \pi \arctan 3}$$