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Calculus help, please!?

0 votes
Find the area of the surface obtained by revolving y=25-x^2 around the y axis on the interval [0,5].

Please explain how you'd do it. Thank you!
asked Apr 13, 2013 in CALCULUS by angel12 Scholar

2 Answers

0 votes

integral subscript 0 superscript 2 y d x space plus integral subscript 2 superscript 5 y d x space equal integral subscript 0 superscript 2 space open parentheses 25 space minus space x to the power of 2 close parentheses d x space plus integral subscript 2 superscript 5 space open parentheses 25 space minus space x to the power of 2 close parentheses d x

integral subscript 0 superscript 2 space open parentheses 25 space minus space x to the power of 2 close parentheses d x space plus integral subscript 2 superscript 5 space open parentheses 25 space minus space x to the power of 2 close parentheses d x space equal open square brackets 25 x space minus x to the power of 3 over 3 close square brackets subscript 0 superscript 2 space plus open square brackets space 25 x space minus space x to the power of 3 over 3 close square brackets subscript 2 superscript 5      (Apply integration)

integral subscript 0 superscript 2 space open parentheses 25 space minus space x to the power of 2 close parentheses d x space plus integral subscript 2 superscript 5 space open parentheses 25 space minus space x to the power of 2 close parentheses d x space equal begin inline style open square brackets 25 open parentheses 2 close parentheses minus begin display style 2 to the power of 3 over 3 end style close square brackets end style space minus open square brackets 25 open parentheses 0 close parentheses minus open parentheses 0 to the power of 3 over 3 close parentheses close square brackets plus begin inline style open square brackets 25 open parentheses 5 close parentheses minus begin display style 5 to the power of 3 over 3 end style close square brackets end style minus begin inline style open square brackets 25 open parentheses 2 close parentheses minus begin display style 2 to the power of 3 over 3 end style close square brackets end style   (Apply limits)

integral subscript 0 superscript 2 space open parentheses 25 space minus space x to the power of 2 close parentheses d x space plus integral subscript 2 superscript 5 space open parentheses 25 space minus space x to the power of 2 close parentheses d x space equal 50 space minus space 8 over 3 minus 0 plus begin inline style open square brackets 125 minus begin display style 125 over 3 end style close square brackets end style minus begin inline style open square brackets 50 minus begin display style 8 over 3 end style close square brackets end style  (Simplify)

integral subscript 0 superscript 2 space open parentheses 25 space minus space x to the power of 2 close parentheses d x space plus integral subscript 2 superscript 5 space open parentheses 25 space minus space x to the power of 2 close parentheses d x space equal 250 over 3      

Therefore The surface area = 83.33sq.units.

answered Apr 13, 2013 by diane Scholar
0 votes

For rotation about y - axis ,the surface area formula is image

  • The curve equation is image

image

image

In this case the interval is [0,5], So a = 0 and b = 5.

  • Our formula becomes image

image

image

image

image

image

Substitute the values in integration.

image

image

image

image

image

image

image

image

image

surface obtained by revolving y = 25 - x^2 around the y axis on the interval [0,5] is

image

 .

answered Jul 4, 2014 by david Expert
edited Jul 4, 2014 by david

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