Find the area of the surface generaled by revolving the

Gwendolyn Willett

Gwendolyn Willett

Answered question

2021-12-16

Find the area of the surface generaled by revolving the curve x=ey+ey2 in the interval 0yln3 about the y-axis

Answer & Explanation

Cheryl King

Cheryl King

Beginner2021-12-17Added 36 answers

Step 1
Given curve is,
x=ey+ey2, 0yln3
Step 2
Required area is,
=0ln3x dx=0ln3(ey+ey2)dy
Step 3
=12[eyey]0ln3 (integral ofex is ex)
=12[(eln3eln3)(e0e0)]
Step 4
=12(eln3eln(3)1) (mlogn=lognm)
=12(313) (eogx=x)
=12×913
=12×83
=43
raefx88y

raefx88y

Beginner2021-12-18Added 26 answers

Step 1
x=eyeyey+ey
x(ey+ey)=eyey
xey+xey=eyey
xey+ey=eyxey
eyxey=xey+ey
(1x)ey=(x+1)ey
(1x)ey(1x)ey=(x+1)ey(1x)ey
eyey=x+11x
ey×1eyx+11x
ey×ey=x+11x
2y=ln(x+11x)
y=12ln(1+x1x)
nick1337

nick1337

Expert2021-12-28Added 777 answers

Step 1
One has;
1) x=eyeyey+ey
Consider the substitution of variables;
u=e2y
then (1) becomes;
x=u1u+1
From this one can isolate u to one side as;
u=x+11x
Upon back substitution one obtains;
2) e2y=x+11x
Taking ln of  both sides of (2) obtains;
2y=ln(x+11x)
or
3) y=12[ln(x+1)ln(1x)]

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