Volume Solution 2

Processing math: 100%

SOLUTION 2: a.) Here are sketches of the base of the solid and the entire solid.

tex2html_wrap_inline125 $\ \ \$

Here are sketches of a square cross-section at

$x$ , together with it's dimensions.

tex2html_wrap_inline125 $\ \ \$

The area of the square cross-section is

$A(x)= (edge)^2 = (\ln x)^2$ . Thus the total volume of this static solid is

$Volume = \int_{1}^{e^2} (\ln x)^2 \ dx$

SOLUTION 2: b.) Here are sketches of the base of the solid and the entire solid.

tex2html_wrap_inline125 $\ \ \$

Here are sketches of a quarter-circular cross-section at

$x$ , together with its radius.

tex2html_wrap_inline125 $\ \ \$

The area of the quarter-circular cross-section is

$A(x)= { 1 \over 4 } \pi r^2 = { 1 \over 4 } \pi (\ln x)^2$ . Thus the total volume of this static solid is

$Volume = \int_{1}^{e^2} { 1 \over 4 } \pi (\ln x)^2 \ dx$

SOLUTION 2: c.) Here are sketches of the base of the solid and the entire solid.

tex2html_wrap_inline125 $\ \ \$

Here are sketches of an isosceles triangular cross-section at

$x$ , together with its dimensions.

tex2html_wrap_inline125 $\ \ \$

The area of the isosceles triangular cross-section is

$A(x)= { 1 \over 2 }(base)(height) = { 1 \over 2 }(\ln x)(4) = 2 \ln x$ . Thus the total volume of this static solid is

$Volume = \int_{1}^{e^2} 2 \ln x \ dx$

Click HERE to return to the list of problems.