Answer:
Volume = [tex]\frac{64}{3}[/tex]
Step-by-step explanation:
Given - Consider the solid S described below. The base of S is the triangular region with vertices (0, 0), (4, 0), and (0, 4). Cross-sections perpendicular to the x-axis are squares.
To find - Find the volume V of this solid.
Solution -
Given that,
The equation of the line with both x-intercept and y-intercept as 4 is -
[tex]\frac{x}{4} + \frac{y}{4} = 1[/tex]
⇒x + y = 4
⇒y = 4 - x
Now,
Volume = [tex]\int\limits^a_b {A(x)} \, dx[/tex]
where
A(x) is the area of general cross-section.
It is given that,
Cross-sections perpendicular to the x-axis are squares.
So,
A(x) = (4 - x)²
As solid lies between x = 0 and x = 4
So,
The Volume becomes
Volume = [tex]\int\limits^4_0 {(4 - x)^{2} } \, dx[/tex]
= [tex]\int\limits^4_0 {[(4)^{2} + (x)^{2} - 8x] } \, dx[/tex]
= [tex]\int\limits^4_0 {[16 + x^{2} - 8x] } \, dx[/tex]
= [tex]{[16 x + \frac{x^{3}}{3} - \frac{8x^{2} }{2} ] } ^4_0[/tex]
= [tex]{[16(4 - 0) + \frac{4^{3}}{3} - \frac{0^{3}}{3} - 4 [4^{2} - 0^{2}] ] }[/tex]
= [tex]{[16(4) + \frac{64}{3} - 0 - 4 [16 - 0] ] }[/tex]
= [tex]{[64 + \frac{64}{3} - 64 ] }[/tex]
= [tex]\frac{64}{3}[/tex]
⇒Volume = [tex]\frac{64}{3}[/tex]
