What is the surface area of the solid created by revolving #f(x)=e^(x^2-x)/(x+1)-e^x# over #x in [0,1]# around the x-axis? Calculus Applications of Definite Integrals Determining the Surface Area of a Solid of Revolution 1 Answer Leland Adriano Alejandro Jan 16, 2016 Area #= 17.01332# square units Explanation: #A=2 pi int_0^1 f(x) ds# #A=2 pi* int_0^1 ((e^(x^2-x))/(x+1)-e^x) (sqrt(1+(((2x^2+x-2)(e^(x^2-x)))/(x+1)^2-e^x)^2) dx# #A=-17.01332# square units Answer link Related questions How do you find the surface area of a solid of revolution? How do you find the surface area of the solid obtained by rotating about the #y#-axis the region... How do you find the surface area of the solid obtained by rotating about the #x#-axis the region... How do you find the surface area of the solid obtained by rotating about the #x#-axis the region... How do you find the surface area of the solid obtained by rotating about the #y#-axis the region... How do you find the surface area of the solid obtained by rotating about the #x#-axis the region... How do you find the surface area of the solid obtained by rotating about the #x#-axis the region... How do you find the surface area of the part of the circular paraboloid #z=x^2+y^2# that lies... How do you determine the surface area of a solid revolved about the x-axis? How do you find the centroid of the quarter circle of radius 1 with center at the origin lying... See all questions in Determining the Surface Area of a Solid of Revolution Impact of this question 1116 views around the world You can reuse this answer Creative Commons License