Question #98070

Find the volume of solid of revolution generated when the area under a curve y*2=x+2 is rotated about the x-axis , between x=0 and x=4.

Expert's answer

Solution To find the volume, we use the formula


V=π∫aby2dxV=\pi \int_{a}^b y^2dx

where y(x) is function is rotated about the x-axis , between x=a and x=b.

For our case we get y=(x+2)/2; a=0 and b=4. Therefore


V=π∫04(x+22)2dx=π(x+2)312∣04=π(18−23)=523πV=\pi \int_{0}^4 (\frac {x+2} {2})^2dx=\pi \frac {(x+2)^3}{12} |_0^4=\pi (18-\frac{2}{3})=\frac {52} {3}\pi



Answer

V=523πV=\frac {52} {3}\pi


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