Question

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. Find the volume V of this solid.

Answer 1

Answer:

Volume = $\frac{64}{3}$

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 -

$\frac{x}{4} + \frac{y}{4} = 1$

⇒x + y = 4

⇒y = 4 - x

Now,

Volume = $\int\limits^a_b {A(x)} \, dx$

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 = $\int\limits^4_0 {(4 - x)^{2} } \, dx$

             = $\int\limits^4_0 {[(4)^{2} + (x)^{2} - 8x] } \, dx$

             = $\int\limits^4_0 {[16 + x^{2} - 8x] } \, dx$

             = ${[16 x + \frac{x^{3}}{3} - \frac{8x^{2} }{2} ] } ^4_0$

             = ${[16(4 - 0) + \frac{4^{3}}{3} - \frac{0^{3}}{3} - 4 [4^{2} - 0^{2}] ] }$

             = ${[16(4) + \frac{64}{3} - 0 - 4 [16 - 0] ] }$

             = ${[64 + \frac{64}{3} - 64 ] }$

             = $\frac{64}{3}$

⇒Volume = $\frac{64}{3}$