The base of a solid is bounded by the curve y= sort (x+2) ,the x-axis and the line x = 1. The cross sections, taken perpendicular to the x-axis, are squares. Find the volume of the solid.
Why did the square cross the x-axis? To find its volume, of course!
Let's tackle this problem step by step. The first thing we need to do is determine the limits of integration. We know that the solid is bounded by the curve y = sqrt(x + 2), the x-axis, and the line x = 1.
To find the limits, we need to find the x-values where the curve, x-axis, and the line x = 1 intersect. For the curve and the x-axis, we have y = sqrt(x + 2) = 0. Solving for x gives us x = -2.
For the line x = 1 and the curve, we have y = sqrt(1 + 2) = sqrt(3). Hence, the limits of integration are -2 to 1.
Now, let's think about the cross section. The problem states that the cross-sections, taken perpendicular to the x-axis, are squares. This means that each cross section has sides of equal length.
To find the length of each side, we need to consider the height of the cross section. The height is given by the function y = sqrt(x + 2). Since the cross section is a square, the length of each side is equal to the height.
Therefore, the area of each cross section is (sqrt(x + 2))^2 = x + 2.
To find the volume of the solid, we need to integrate the area function from -2 to 1:
∫(x + 2) dx = [x^2/2 + 2x] from -2 to 1
Plugging in the limits, we get:
(1^2/2 + 2(1)) - (-2^2/2 + 2(-2))
= (1/2 + 2) - (4/2 - 4)
= (1/2 + 2) - (2 - 4)
= 1/2 + 2 - 2 + 4
= 4.5
So, the volume of the solid is 4.5 units^3. Ta-da!
To find the volume of the solid, we need to integrate the area of each square cross section along the x-axis.
First, let's find the limits of integration. The curve y = sqrt(x+2) intersects the x-axis when y = 0. So, we can solve for x:
0 = sqrt(x+2)
0^2 = (sqrt(x+2))^2
0 = x + 2
x = -2
The curve y = sqrt(x+2) also intersects the vertical line x = 1. So, the limits of integration will be -2 to 1.
Next, let's find the side length of each square cross section. Since the cross sections are squares, the side length will be equal to the height of each cross section. In this case, the height is given by the curve y = sqrt(x+2).
So, the area of each cross section will be (side length)^2 = (sqrt(x+2))^2 = x + 2.
Finally, we integrate the area of each cross section over the interval [-2, 1] to find the volume:
Volume = ∫[from -2 to 1] (x + 2) dx
Integrating this expression gives:
Volume = [(1/2)x^2 + 2x] evaluated from -2 to 1
Volume = (1/2(1)^2 + 2(1)) - (1/2(-2)^2 + 2(-2))
Volume = (1/2 + 2) - (1/2 + 4)
Volume = 1/2 - 2
Volume = -3/2
Therefore, the volume of the solid is -3/2 cubic units.
To find the volume of the solid, we can use integration. Since the cross sections are squares, we can find the area of each square cross section and integrate it.
The curve y = sqrt(x + 2) intersects the x-axis at the point where y = 0. Substituting y = 0 into the equation, we get:
0 = sqrt(x + 2)
Squaring both sides, we get:
0 = x + 2
Solving for x, we find that x = -2. This means that the solid extends from x = -2 to x = 1.
To find the side length of the square cross section at any given x value, we evaluate the difference in y-values between the curve and the x-axis. The side length of the square will be 2 times this difference since the cross section is symmetrical. Therefore, the side length of the square is 2 * sqrt(x + 2).
Now, we can set up the integral for the volume:
V = ∫[from -2 to 1] (2 * sqrt(x + 2))^2 dx
Simplifying the expression, we get:
V = 4 * ∫[from -2 to 1] (x + 2) dx
Evaluating the integral, we get:
V = 4 * [x^2/2 + 2x] [from -2 to 1]
V = 4 * [(1^2/2 + 2(1)) - ((-2)^2/2 + 2(-2))]
V = 4 * [1/2 + 2 - 2 + 4]
V = 4 * (5/2)
V = 10
So, the volume of the solid is 10 cubic units.
well, clearly each cross-section has a base of side length y, so its area is y^2
The volume is thus
∫[-2,1] y^2 dx
= ∫[-2,1] (x+2) dx
I'm sure you can take it from here, no?