# Base of a certain solid

1. Mar 17, 2005

The base of a certain solid is the triangle with vertices at (-8,4) ,(4,4), and the origin. Cross-sections perpendicular to the y-axis are squares.

Then the volume of the solid is __________

i have no clue or dont even know how to start this problem. Well the first thing i did was just graphed the three points they gave me, but that's about it. can someone lend me a hand?

2. Mar 17, 2005

### Gamma

The hight of this solid varies along the y axis. Cosider an element of size dy at a distance y from the origin along the y axis. Find the width of this element using the given information in terms of y. What is the height of the element. (same as the width since cross section is squre). So now you can find the volume of the element.
call this dv
Intergrate dv from y=0 to y= y1 where y1 is the distance from origin to the base of the triangle.

3. Mar 18, 2005

### xanthym

From problem statement:
{Base Vertices (0,0) & (-8,4)} ⇒ {Bounding Base Line} = {y = (-1/2)*x} = {x = (-2)*y}
{Base Vertices (0,0) & (4,4)} ⇒ {Bounding Base Line} = {y = x} = {x = y}
{Base Vertices (-8,4) & (4,4)} ⇒ {Bounding Base Line} = {y = 4}

{Cross-section Area ⊥ y-axis} = {Width}*{Height} =
= {y - (-2)*y}*{y - (-2)*y} = ::: (Square cross-section ⇒ height=width)
= {3y}*{3y} =
= 9*y2

{Differential Volume} = dV = {Cross-section Area}*dy = {9*y2}*dy

$$\ \ \ \ (Volume) \ = \ \int_{0}^{4} 9y^{2} dy \ = \ \left [ 3y^{3} \right ]_{0}^{4}$$

$$\ \ \ \color{red} (Volume) \ = \ (192)$$

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Last edited: Mar 18, 2005
4. Mar 18, 2005