Answer to Question #257707 in Civil and Environmental Engineering for toto

Question #257707

Find the volume generated under the curve y=x^2 + 1 and the x-axis between x=0 and x=3, rotated about the y-axis 


1
Expert's answer
2021-10-30T11:19:53-0400

Solution To find the volume, we use the formula



"V=\\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=\\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=\\frac {52} {3}\\pi"

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