anonymous
  • anonymous
find the region of the solid formed by rotating the area bounded by y=3, y=sqrt(x), and the x-axis. Rotate around x=10.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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ash2326
  • ash2326
we have the curve y=sqrt (x)|dw:1327469538153:dw|
ash2326
  • ash2326
if we have to calculate area bounded, it'll be between the curves x=y^2 and x=10 take a rectangle of length 10-y^2 and thickness dy, y will vary from y=3 to y= root 10 \[\int\limits_{3}^{\sqrt{10}}( 10-y^2) dy\] now the area is rotated about x=10 so the volume of a disc= \(\pi*(10-y^2)*dy\) radius is 10-y^2 thickness is dy y will vary from y=3 to y= \(\sqrt 10\) Volume =\(\int\limits_{3}^{\sqrt{10}} \pi* (10-y^2)^2dy\) solving the integral, you'll get volume =0.1712 http://www.wolframalpha.com/input/?i=integrate+%28pi*%2810-y%5E2%29%5E2%29dy+from+y%3D3+to+y%3D+10%5E%281%2F2%29

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