The base of a certain solid is the triangle with vertices at (−6,3), (3,3), and the origin. Cross-sections perpendicular to the y-axis are squares. Then the volume of the solid?
Think of the solid as a stack of thin squares. At a distance y from the origin, the square has side 3y. So, add them all up and the volume is
∫[0,3] (3y)^2 dy
To find the volume of the solid, we can use the method of integration.
Let's start by visualizing the situation. We are given that the base of the solid is a triangle with vertices at (-6,3), (3,3), and the origin. The base lies on the x-axis and has a length of 9 units. The cross-sections of the solid, perpendicular to the y-axis, are squares. This means that for each y-value within the range of the base triangle, we have a square cross-section whose side length is equal to the difference between the x-coordinates of the vertices of the base triangle at that y-value.
To determine the range of y-values for the cross-sections, we look at the y-coordinates of the vertices of the base triangle. In this case, the y-coordinate is always 3. So, the range of y-values is from y = 0 to y = 3.
Now, let's express the volume of the solid as an integral. We integrate the area of each square cross-section over the range of y-values.
The area of each square cross-section can be calculated as the square of the difference in x-coordinates at each y-value. In this case, the difference in x-coordinates is a constant value of 9.
Therefore, the integral that represents the volume of the solid is:
V = ∫[0,3] (9)^2 dy
Now, we can solve this integral to find the volume. Integrating (9)^2 dy gives us:
V = 81y |[0,3]
V = 81(3) - 81(0)
V = 243 cubic units
Hence, the volume of the solid is 243 cubic units.
To find the volume of the solid, we first need to determine the dimensions of the cross-sections perpendicular to the y-axis.
The base of the solid is given by the triangle with vertices at (-6,3), (3,3), and the origin (0,0).
The length of the base can be found by calculating the distance between the points (-6,3) and (3,3) on the x-axis. The distance formula is:
Distance = √((x2 - x1)^2 + (y2 - y1)^2)
For the given points, the length of the base is:
Distance = √((3 - (-6))^2 + (3 - 3)^2)
= √(9^2 + 0^2)
= √(81 + 0)
= √81
= 9
Since the cross-sections perpendicular to the y-axis are squares, the width and height of each cross-section will also be 9 units.
Therefore, the volume of the solid can be calculated by multiplying the area of the base by the length of the solid. Since the base is a triangle, its area can be found by using the formula for the area of a triangle:
Area = (1/2) * base * height
In this case, the base of the triangle is 9 units and the height can be calculated by finding the distance between the points (0,0) and (0,3) on the y-axis:
Height = 3 - 0
= 3
Substituting the values into the formula, we have:
Area = (1/2) * 9 * 3
= (1/2) * 27
= 13.5
The length of the solid is not provided in the question. If you have the length, you can multiply it by the area of the base to find the volume of the solid.