The base of a solid is a region bounded by the curve (x^2/8^2)+(y^2/4^2)=1 (an ellipse with the major and minor axes of lengths 16 and 8 respectively). Find the volume of the solid if every cross section by a plane perpendicular to the major axis (x-axis) has the shape of an isosceles triangle with height equal to 1/4 the length of the base.
This is just like the other one The base of each triangle has length 2x,
making the area 1/2 bh = 1/2 (2x)(x/2) = 1/2 x^2
Using the symmetry of the figure,
v = 2∫[0,8] 1/2 x^2 dx
same oops as before.
v = 2∫[0,8] 1/2 y^2 dx = v = 2∫[0,8] 1/2 (16 - x^2/4) dx
To find the volume of the solid, we need to determine the area of the cross-section first.
Let's consider a cross-section of the solid formed by a plane perpendicular to the major axis (x-axis). This cross-section is in the shape of an isosceles triangle, with the base parallel to the x-axis.
The length of the base of the isosceles triangle is given by the equation of the ellipse as 2a, where a is the semi-major axis (length of major axis/2). In this case, a = 16/2 = 8.
The height of the isosceles triangle is 1/4 times the length of the base. So, the height h of the triangle is (1/4) * 2a = a/2 = 8/2 = 4.
Now we can calculate the area of the cross-section:
Area = (1/2) * base * height
= (1/2) * 2a * (a/2)
= a^2/2
= 8^2/2
= 64/2
= 32
Therefore, the area of the cross-section is 32 square units.
To find the volume of the solid, we need to integrate the area of the cross-section along the length of the major axis.
The length of the major axis is 2a = 2 * 8 = 16.
Therefore, the volume V of the solid can be calculated as:
V = ∫[0, 16] Area dx
V = ∫[0, 16] 32 dx
= 32 * [x]ᴏ¹⁶₀
= 32 * (16 - 0)
= 512 units³
Hence, the volume of the solid is 512 cubic units.
To find the volume of the solid, we need to integrate the areas of the cross sections as we move along the x-axis. Since each cross section is an isosceles triangle, we need to find the length of the base of each triangle.
Let's first consider a cross section at a specific x-coordinate, say x. We can treat this cross section as a simple triangle.
Since the base of the triangle is defined by the ellipse, we need to find the corresponding y-coordinates that lie on the curve for the given x-coordinate.
The equation of the ellipse is given by:
(x^2/8^2) + (y^2/4^2) = 1
Simplifying this equation, we have:
(x^2/64) + (y^2/16) = 1
Rearranging, we get:
y^2/16 = 1 - (x^2/64)
Now, multiply both sides by 16 to isolate y^2:
y^2 = 16 - (x^2/4)
Taking the square root of both sides, we get:
y = ±√(16 - (x^2/4))
Since we are looking for the positive y-values to form the upper part of the ellipse, we can write the equation as:
y = √(16 - (x^2/4))
Now, we know that the height of each triangle is equal to 1/4 the length of its base. Thus, the height (h) is given by:
h = (1/4) * (2 * √(16 - (x^2/4)))
The base of each triangle is given by the length between the two corresponding y-values on the curve. So, the base (b) is given by:
b = 2 * √(16 - (x^2/4))
Now, the area (A) of the isosceles triangle is given by:
A = (1/2) * b * h
Substituting the values for b and h, we get:
A = (1/2) * (2 * √(16 - (x^2/4))) * ((1/4) * (2 * √(16 - (x^2/4))))
Simplifying further, we have:
A = √(16 - (x^2/4))
To find the volume (V) of the solid, we integrate the areas of the cross sections from one end of the ellipse to the other along the x-axis. Since the ellipse extends from -a to a (the major axis), the volume is given by:
V = ∫[from -a to a] √(16 - (x^2/4)) dx
Evaluating this integral will give us the volume of the solid.