Question 1082572
*[illustration cf9.JPG].
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{{{y=e^(2x)+4}}}
{{{e^(2x)=y-4}}}
{{{2x=ln(y-4)}}}
{{{x=ln(y-4)/2}}}
and
{{{x=2}}}
So,
{{{A(y)=pi((2)^2-(ln(y-4)/2)^2)}}}
{{{A(y)=pi(4-(ln(y-4))^2/4)}}}
and
{{{V=int((A(y)),dy,y=5,e^(4)+4)}}}

{{{V=int((pi(4-(ln(y-4))^2/4)),dy,y=5,e^(4)+4)}}}
{{{V=(1/4)(14y-(y-4)ln(y-4)^2+2(y-4)ln(y-4)+8)+C}}}
{{{V=(1/4)(14(e^4+4)-(e^4+4-4)ln(e^4+4-4)^2+2(e^4+4-4)ln(e^4+4-4)+8)-
     (1/4)(14(5)-(5-4)ln(5-4)^2+2(5-4)ln(5-4)+8)}}}
{{{V=(1/4)(14(e^4+4)-(e^4)ln(e^4)^2+2(e^4)ln(e^4)+8)-
     (1/4)(14(5)-(1)ln(1)^2+2(1)ln(1)+8)}}}
{{{V=(1/4)(14(e^4)+56-16e^4+8e^4+8)-
     (1/4)(70+8)}}}
{{{V=(1/4)(6(e^4)+64)-
     (1/4)(78)}}}
{{{V=(1/4)(6(e^4)-14)}}}
{{{highlight(V=(1/2)(3(e^4)-7))}}}