Question

The solid S has a base region B defined by the curves y = 5x − x 2 and y = x. (A) Find the volume of S if the cross-sections through S perpendicular to the x-axis are squares with an edge along the base. (B) Find the volume of S if the cross-sections through S perpendicular to the x-axis are equilateral triangles with an edge along the base. 3 (C) Find the volume of S if the cross-sections through S parallel to the x-axis are semicircles with their diameter along the base. (D) Find the volume of S if the cross-sections through S parallel to the x-axis are isosceles right triangles with a leg along the base.

Answer 1

Answer:

a) The volume of S is 34.13

b) The volume of S is 14.8

c) The volume of S is 5.17

d) The volume of S is 11.33

Step-by-step explanation:

a) The cross section area is equal to:

$A=a^{2} =((5x-x^{2})-x)^{2} =(4x-x^{2} )^{2}$

The volume of S is equal to:

$Vol_{S} =\int\limits^4_0 {A(x)} \, dx =\int\limits^4_0 {(4x-x^{2})^{2} } \, dx =34.13$

b) The cross section area is equal to:

$A=\frac{a^{2}\sqrt{3} }{4} =\frac{\sqrt{3} }{4} ((5x-x^{2} )-x)^{2} =\frac{\sqrt{3} }{4} (4x-x^{2} )^{2}$

The volume of S is equal to:

$Vol_{S} =\int\limits^4_0 {A(x)} \, dx =\frac{\sqrt{3} }{4} \int\limits^4_0 {(4x-x^{2})^{2} } \, dx =14.8$

c)

$y=5x-x^{2} \\\frac{dy}{dx} =0\\5x-x^{2} =0\\x=5/2\\y(5/2)=25/4\\y=5x-x^{2} \\x^{2} -5x+y=0\\x=\frac{5+-\sqrt{25-4y} }{2}$

The cross section area is equal to:

$A_{1} =\frac{1}{2} \pi r_{1}^{2} =\frac{1}{2} \pi (\frac{1}{2} (\frac{5+\sqrt{25-4y} }{2} -\frac{5-\sqrt{25-4y} }{2} ))^{2} =\frac{1}{8} \pi (25-4y)\\A_{2} =\frac{1}{2} \pi r_{2}^{2}=\frac{1}{2}\pi (\frac{1}{2} (y-\frac{5-\sqrt{25-4y} }{2} ))^{2} =\frac{1}{32} \pi (2y-5+\sqrt{25-4y} )^{2}$

The volume of S is equal to:

$Vol_{S} =\int\limits^a_b {A_{1}(y) } \, dy+\int\limits^4_0 {A_{2}(y) } \, dy ,where-a=25/4,b=4\\Vol_{S} =\int\limits^a_b {\frac{1}{8}\pi (25-4y)} \, dy +\int\limits^a_b {\frac{1}{32}\pi (2y-5+\sqrt{25-4y} )^{2} } \, dy =5.17$

d) The cross section area is:

$A_{1} =\frac{1}{2}ab=\frac{1}{2} a^{2} =\frac{1}{2} (\frac{5+\sqrt{25-4y} }{2}-\frac{5-\sqrt{25-4y}}{2} )^{2} =\frac{1}{2} (25-4y)\\A_{1}=\frac{1}{2}ab=\frac{1}{2} a^{2} =\frac{1}{2}(y-\frac{5-\sqrt{25-4y}}{2}} )^{2} =\frac{1}{8} (2y-5+\sqrt{25-4y}})^{2}$

The volume of S is equal to:

$Vol_{S} =\int\limits^a_b {A_{1}(y) } \, dy +\int\limits^4_0 {A_{2}(y) } \, dy ,where-a=25/4,b=4\\Vol_{S}=\int\limits^a_b {\frac{1}{2}(25-4y) } \, dy +\int\limits^4_0 {\frac{1}{8}(2y-5+\sqrt{25-4y})^{2} } \, dy =11.33$