Questions: Problem. 23: Use a double integral to find the volume of the solid shown in the figure. ? cubic units Problem. 24: Find the volume of the solid bounded above by the surface z=f(x, y) and below by the plane region R. f(x, y)=3y/(1+x^2) ; R is the region bounded by y=0, y=sqrt(x), x=16 ? cubic units Problem. 25: The price of land in a rectangular region R of a city is approximated by the function p(x, y)=150-13(x-1/2)^2-13(y-1)^2, 0 ≤ x ≤ 1, 0 ≤ y ≤ 2 where p(x, y) is the price of land in dollars per square foot and x and y are measured in miles. What is the average price of land per square foot in R? (Round your answer to the nearest dollar.) 1575

Problem. 23: Use a double integral to find the volume of the solid shown in the figure.
?
cubic units

Problem. 24: Find the volume of the solid bounded above by the surface z=f(x, y) and below by the plane region R.
f(x, y)=3y/(1+x^2) ; R is the region bounded by y=0, y=sqrt(x), x=16
?
cubic units

Problem. 25: The price of land in a rectangular region R of a city is approximated by the function
p(x, y)=150-13(x-1/2)^2-13(y-1)^2, 0 ≤ x ≤ 1, 0 ≤ y ≤ 2
where p(x, y) is the price of land in dollars per square foot and x and y are measured in miles. What is the average price of land per square foot in R? (Round your answer to the nearest dollar.)
1575

Transcript text: Problem. 23 : Use a double integral to find the volume of the solid shown in the figure. ? cubic units Problem. 24 : Find the volume of the solid bounded above by the surface $z=f(x, y)$ and below by the plane region $R$. $f(x, y)=\frac{3 y}{1+x^{2}} ; \quad R$ is the region bounded by $y=0, y=\sqrt{x}, x=16$ ? cubic units Problem. 25 : The price of land in a rectangular region $R$ of a city is approximated by the function \[ p(x, y)=150-13\left(x-\frac{1}{2}\right)^{2}-13(y-1)^{2}, \quad 0 \leq x \leq 1, \quad 0 \leq y \leq 2 \] where $p(x, y)$ is the price of land in dollars per square foot and $x$ and $y$ are measured in miles. What is the average price of land per square foot in $R$ ? (Round your answer to the nearest dollar.) \$ 1575

Solution

Solution Steps

Step 1: Set up the double integral for the volume

To find the volume of the solid using a double integral, we need to set up the integral with the given boundaries. The solid is bounded by $ z = 5 $, $ y = 4 - x $, and $ y = x $.

Step 2: Determine the limits of integration

The region in the xy-plane is bounded by $ y = x $ and $ y = 4 - x $. The limits for $ x $ are from 0 to 2 (where $ y = x $ intersects $ y = 4 - x $). For each $ x $, $ y $ ranges from $ x $ to $ 4 - x $.

Step 3: Set up and evaluate the double integral

The volume $ V $ is given by: \[ V = \int_{0}^{2} \int_{x}^{4-x} 5 \, dy \, dx \]

Evaluate the inner integral with respect to $ y $: \[ \int_{x}^{4-x} 5 \, dy = 5[y]_{x}^{4-x} = 5(4-x - x) = 5(4 - 2x) = 20 - 10x \]

Now, evaluate the outer integral with respect to $ x $: \[ V = \int_{0}^{2} (20 - 10x) \, dx \]

\[ V = \left[ 20x - 5x^2 \right]_{0}^{2} = (20(2) - 5(2)^2) - (20(0) - 5(0)^2) \]

\[ V = (40 - 20) - 0 = 20 \]