What is the Volume of a Solid with a Triangular Base and Square Cross-Sections?

[ProBasket]

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 don't 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?

 

[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.

 

[xanthym]

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 don't 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?

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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[ProBasket]

thanks so much