An ellipsoid has radii with lengths of 5 5, 5 5, and 8 8. A portion the size of a hemisphere with a radius of 3 3 is removed form the ellipsoid. What is the remaining volume of the ellipsoid?

1 Answer
Feb 22, 2018

see a solution step below;

Explanation:

Please note that we are dealing with an Ellipsoid and Hemisphere..

The formula for the volume of an ellipsoid is given below;

"Volume of an ellipsoid" rArr V_e = 4/3piabcVolume of an ellipsoidVe=43πabc

The formula for the volume of a hemisphere is given below;

"Volume of a hemisphere" rArr V_h = 2/3pir^3Volume of a hemisphereVh=23πr3

Since we are removing a portion of a Hemisphere from an Ellipsoid, it means that we need to subtract the volume of a hemisphere from the ellipsoid which is given below;

"Remaining volume of an ellipsoid" color(white)x R_(ve) = V_e - V_hRemaining volume of an ellipsoidxRve=VeVh

R_(ve) = V_e - V_hRve=VeVh

Where;

R_(ve) = "Remaining volume of an ellipsoid"Rve=Remaining volume of an ellipsoid

V_e = "Volume of an ellipsoid"Ve=Volume of an ellipsoid

V_h = "Volume of a hemisphere"Vh=Volume of a hemisphere

a, b, c = "lengths"a,b,c=lengths

r = "radius"r=radius

R_(ve) = V_e - V_hRve=VeVh

V_e = 4/3piabcVe=43πabc

V_h = 2/3pir^3Vh=23πr3

a, b, c = 5, 5, 8 "respectively"a,b,c=5,5,8respectively

r = 3r=3

pi = 3.142π=3.142

Hence substituting the values into the formula;

R_(ve) = V_e - V_hRve=VeVh

R_(ve) = 4/3pi(5 xx 5 xx 8)- 2/3pi(3)^3Rve=43π(5×5×8)23π(3)3

R_(ve) = 4/3pi(200)- 2/3pi(27)Rve=43π(200)23π(27)

R_(ve) = (4(200))/3 pi - (2(27))/3 piRve=4(200)3π2(27)3π

R_(ve) = 800/3 pi - 54/3 piRve=8003π543π

R_(ve) = (800 - 54)/3 piRve=800543π

R_(ve) = 746/3 piRve=7463π

R_(ve) = 248.67 piRve=248.67π

R_(ve) = 248.67 xx 3.142Rve=248.67×3.142

R_(ve) = 781.31cm^3Rve=781.31cm3

Therefore the remaining volume of the ellipsoid is 781.31cm^3781.31cm3