Find the volume of the solid whose base is the circle x^2+y^2=25 and the cross sections perpendicular to the x-axis are triangles whose height and base are equal.
Find the area of the vertical cross section A at the level x=1.
Well, that's an interesting shape we've got here! So, let's break it down step by step.
First, we need to figure out the limits of integration in order to find the bounds for x. The equation of the circle is x^2 + y^2 = 25, which is a circle with radius 5. We know that the base of our solid is this circle, so the bounds for x would be -5 to 5.
Now, let's talk about the cross sections. We're told that they are triangles, with the height and base being equal. So, basically, we've got isosceles triangles. Let's call the height and base of each triangle h.
To find h, we can look at the equation of the circle x^2 + y^2 = 25. At any given x-value, the y-values that satisfy this equation will give us the height of the corresponding triangle.
So, for the cross section at x=1, we can plug in x=1 into the equation of the circle: 1^2 + y^2 = 25. Solving for y, we get y = ±√24. Since we're finding the height, we'll take the positive value, so h = √24.
Now that we have the height of the cross section at x=1, we can find the area of the triangle cross section. The formula for the area of an isosceles triangle is A = (1/2) * base * height. Since the base and height are equal in our case, the area would be A = (1/2) * h^2.
Substituting the value of h we found earlier, we get A = (1/2) * (√24)^2 = (1/2) * 24 = 12.
So, the area of the vertical cross section A at the level x=1 is 12.
To find the volume of the solid, we need to integrate the area of each cross section over the range of x-values.
First, let's find the height of the triangles at each x-value by finding the y-coordinate of the points on the circle.
Given the equation of the circle: x^2 + y^2 = 25,
Solving for y, we get:
y = sqrt(25 - x^2)
Since the height and base of the triangles are equal, the height of each triangle is also equal to sqrt(25 - x^2).
Now, let's find the area of each cross section at each x-value.
The area of a triangle is given by the formula: A = (1/2) * base * height.
In this case, both the base and the height are equal to sqrt(25 - x^2).
Therefore, the area of each cross section is A = (1/2) * (sqrt(25 - x^2)) * (sqrt(25 - x^2)) = (1/2) * (25 - x^2).
To find the total volume, integrate the area of each cross section from x = -5 to x = 5:
V = ∫[from -5 to 5] (1/2) * (25 - x^2) dx
To find the area of the vertical cross section A at the level x = 1, substitute x = 1 into the equation (1/2) * (25 - x^2):
A = (1/2) * (25 - 1^2)
= (1/2) * (25 - 1)
= (1/2) * 24
= 12 square units.
Therefore, the area of the vertical cross section A at the level x = 1 is 12 square units.
To find the volume of the solid, we first need to determine the limits of integration for the variable x. Looking at the base, which is the circle x^2 + y^2 = 25, we can solve for y in terms of x:
y = √(25 - x^2) and y = -√(25 - x^2)
The height and base of the cross sections are equal, so the triangles formed will be isosceles triangles with two equal sides.
Let's consider an arbitrary vertical cross section at a specific x value, say x = a. The height and base of this cross section will be equal to the distance between the upper and lower points of the circle at x = a.
For x = a, the coordinates of the points on the circle are (a, √(25 - a^2)) and (a, -√(25 - a^2)). The distance between these points is:
2√(25 - a^2)
This distance represents both the height and base of the cross section. So, the area of this cross section A at x = a is:
A = (1/2) * base * height = (1/2) * 2√(25 - a^2) * 2√(25 - a^2) = 2(25 - a^2)
To find the volume of the solid, we'll integrate the area of these cross sections over the range of x, from the values where the circle intersects the x-axis.
The circle intersects the x-axis at x = -5 and x = 5. So, the limits of integration for x are -5 to 5.
V = ∫[from -5 to 5] 2(25 - a^2) da
Integrating, we get:
V = 2 ∫[from -5 to 5] (25a - a^3) da
Now, let's integrate term by term:
V = 2 ( (25/2)*a^2 - (1/4)*a^4 ) evaluated from -5 to 5
Substituting the limits, we have:
V = 2 ( (25/2)*(5^2) - (1/4)*(5^4) ) - 2 ( (25/2)*(-5^2) - (1/4)*(-5^4) )
V = 2 ( (25/2)*25 - (1/4)*625 ) - 2 ( (25/2)*25 - (1/4)*625 )
V = 2 ( (625/2) - (625/4) ) - 2 ( (625/2) - (625/4) )
V = 250 - 125
V = 125 cubic units
So, the volume of the solid is 125 cubic units.
To find the area of the vertical cross section A at the level x = 1, we substitute x = 1 into the expression for the area:
A = 2(25 - 1^2) = 2(24) = 48 square units
Therefore, the area of the vertical cross section A at the level x = 1 is 48 square units.
The triangle with base at x has base with length 2y where y is on the circle, so the area A of each triangle is
1/2 (2y)(2y) = 2y^2 = 2(25-x^2)
The volume is just the integral of that from -5 to 5.