Surface Area of a revolution
$begingroup$
So I had a question involving Surface Area.
Find the surface area generated by revolving the curve $x = y^3$ about the $y$-axis for $0 leq y leq sqrt[4]{11}$.
(a) $23piquad$ (b) $37piquad$ (c) $46piquad$ (d) $62piquad$ (e) $73piquad$ (f) None of these
This was my Attempt but I got stuck with trying to integrate the stuff under the radical.
calculus integration area surfaces
asked Dec 17 '18 at 1:40
krauser126krauser126
475
$endgroup$
add a comment |
$begingroup$
So I had a question involving Surface Area.
Find the surface area generated by revolving the curve $x = y^3$ about the $y$-axis for $0 leq y leq sqrt[4]{11}$.
(a) $23piquad$ (b) $37piquad$ (c) $46piquad$ (d) $62piquad$ (e) $73piquad$ (f) None of these
This was my Attempt but I got stuck with trying to integrate the stuff under the radical.
calculus integration area surfaces
asked Dec 17 '18 at 1:40
krauser126krauser126
475
$endgroup$
add a comment |
$begingroup$
So I had a question involving Surface Area.
Find the surface area generated by revolving the curve $x = y^3$ about the $y$-axis for $0 leq y leq sqrt[4]{11}$.
(a) $23piquad$ (b) $37piquad$ (c) $46piquad$ (d) $62piquad$ (e) $73piquad$ (f) None of these
This was my Attempt but I got stuck with trying to integrate the stuff under the radical.
calculus integration area surfaces
asked Dec 17 '18 at 1:40
krauser126krauser126
475
$endgroup$
So I had a question involving Surface Area.
Find the surface area generated by revolving the curve $x = y^3$ about the $y$-axis for $0 leq y leq sqrt[4]{11}$.
(a) $23piquad$ (b) $37piquad$ (c) $46piquad$ (d) $62piquad$ (e) $73piquad$ (f) None of these
This was my Attempt but I got stuck with trying to integrate the stuff under the radical.
calculus integration area surfaces
calculus integration area surfaces
asked Dec 17 '18 at 1:40
krauser126krauser126
475
asked Dec 17 '18 at 1:40
krauser126krauser126
475
edited Dec 17 '18 at 1:49
krauser126
asked Dec 17 '18 at 1:40
krauser126krauser126
475
asked Dec 17 '18 at 1:40
krauser126krauser126
475
asked Dec 17 '18 at 1:40
krauser126krauser126
475
475
add a comment |
add a comment |
1 Answer
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Hint: $int y^3(9(y^4+frac19))^{frac12}operatorname dy=frac32(y^4+frac19)^{frac 32}+C$.
answered Dec 17 '18 at 1:53
Chris CusterChris Custer
14.2k3827
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add a comment |
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1 Answer
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1 Answer
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$begingroup$
Hint: $int y^3(9(y^4+frac19))^{frac12}operatorname dy=frac32(y^4+frac19)^{frac 32}+C$.
answered Dec 17 '18 at 1:53
Chris CusterChris Custer
14.2k3827
$endgroup$
add a comment |
$begingroup$
Hint: $int y^3(9(y^4+frac19))^{frac12}operatorname dy=frac32(y^4+frac19)^{frac 32}+C$.
answered Dec 17 '18 at 1:53
Chris CusterChris Custer
14.2k3827
$endgroup$
add a comment |
$begingroup$
Hint: $int y^3(9(y^4+frac19))^{frac12}operatorname dy=frac32(y^4+frac19)^{frac 32}+C$.
answered Dec 17 '18 at 1:53
Chris CusterChris Custer
14.2k3827
$endgroup$
Hint: $int y^3(9(y^4+frac19))^{frac12}operatorname dy=frac32(y^4+frac19)^{frac 32}+C$.
answered Dec 17 '18 at 1:53
Chris CusterChris Custer
14.2k3827
answered Dec 17 '18 at 1:53
Chris CusterChris Custer
14.2k3827
answered Dec 17 '18 at 1:53
Chris CusterChris Custer
14.2k3827
answered Dec 17 '18 at 1:53
Chris CusterChris Custer
14.2k3827
14.2k3827
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