Surface Area of the solid of revolution by rotating curve:?

Question

a)y=cos(3x) from x=0 to x=pi/6 about the x-axis

b)x=(1/9)(y^2 + 6)^(3/2) from y=1 to y=6 about the x-axis

c)x=sqrt(2y-y^2) from y=10/16 to y=17/16 about the y-axis

how do you go about doing these types of questions

Answer 1

You should know from your notes or textbook that the formula for the surface area of the solid of revolution obtained by rotating the curve y=f(x) between x=a and x=b about the x-axis is

S = 2pi int_{a}^{b} f(x)sqrt(1+f '(x)^2) dx
with an analogous equation if we rotate the curve x=g(y) between y=c and y=d about the y-axis (replace limits a and b by c and d, function f(x) by g(y) and integrate with respect to y).

a) Here a=0, b = pi/6 and f(x) = cos(3x), so f '(x) = -3sin(3x).
Now
1 + f '(x)^2 = 1 + 9sin^2(3x)
and the surface area formula gives

S = 2pi int_{0}^{pi/6} cos(3x)[1+9sin^2(3x)]^(1/2) dx.

To work out this integral, try the substitution
u = 3sin(3x), then du/dx = 9cos(3x) so 1/9 du = cos(3x) dx
and the limits of integration become
u =3sin(0) =0 and u=3sin(pi/2) = 3.
In terms of u the integral becomes

S = 2pi int_{0}^{3} 1/9[1+u^2]^(1/2) du
= 2pi/9 [u/2*(1+u^2)^(1/2)+
1/2*ln(u+(u^2+1)^(1/2))]_{0}^{3}

(just looked up the integral from a standard table)

= pi/9 [3*(1+9)^(1/2)+
ln(3+(9+1)^(1/2)) - 0 ]
= pi/9 [3sqrt(10) +ln(3+sqrt(10))].

b) Here we're still rotating about the x-axis so we need to change the given function and limits into the form y=f(x) with x= a and b.

Making y the subject of the expression gives
(y^2+6)^(3/2) = 9x
(y^2+6) = (9x)^(2/3) [raising both sides to the power (2/3)]
y = sqrt((9x)^(2/3) -6).

Also, when y= 1, x= 1/9*7^(3/2)
and when y=6, x = 1/9*42^(3/2).

Now f(x) = ((9x)^(2/3)-6)^(1/2)
so f '(x) = 1/2*((9x)^(2/3)-6)^(-1/2)*9^(2/3)*(2/3*x^(-1/3))
= 1/3*9^(2/3)*((9x)^(2/3)-6)^(-1/2)*x^(-1/3)
and
f '(x)^2 = 1/9*9^(4/3)*((9x)^(2/3)-6)^(-1)*x^(-2/3)
so 1+ f '(x)^2
= 1 + 9^(1/3)*((9x)^(2/3)-6)^(-1)*x^(-2/3)
and the surface area is given by the integral

S = 2pi int_{1/9*7^(3/2)}^{1/9*42^(3/2)}
((9x)^(2/3)-6)^(1/2)*sqrt[1 + 9^(1/3)*((9x)^(2/3)-6)^(-1)*x^(-2/3)] dx
which looks pretty gruesome but the integrand (expression under the integral) does simplify to

sqrt[(9x)^(2/3)-6+9^(1/3)x^(-2/3)] and the term in [ ] is a perfect square!

Check: [(9x)^(1/3) - (3x)^(-1/3)]^2 = (9x)^(2/3) - 2.3 + (3x)^(-2/3) = (9x)^(2/3) - 6 + 9^(1/3)x^(-2/3).

Integrating gives
9^(1/3)*3/4*x^(4/3) - 3^(-1/3)*3/2*x^(2/3)
= 3^(2/3)/4*[3x^(4/3) - 2x^(2/3)]

When x= 1/9*7^(3/2), x^(2/3) = 3^(-4/3)*7, x^(4/3) = 3^(-8/3)*7^2
and when x = 1/9*42^(3/2), x^(2/3) = 3^(-4/3)*42 and x^(4/3) = 3^(-8/3)*(42)^2.

FINALLY we have

S = 2pi*3^(2/3)/4*[{3*3^(-8/3)*(42)^2 - 2*3^(-4/3)*42}
- {3*3^(-8/3)*7^2 - 2*3^(-4/3)*7}]

= pi/2*3^(2/3)*[3^(-5/3){42^2-7^2} - 2*3^(-4/3){42-7}].

Answer 2

Draw a sketch of the curve that is to be revolved around the axis of revolution. Pick out some tiny piece of the curve and draw a line segment from this piece to the axis of revolution. Let r be the length of this line segment. Then revolving the tiny piece of curve about the axis of revolution contributes an amount dA of surface area given by

dA = 2πr ds

where ds is the length of the tiny piece of curve. If the curve is given as y as a function of x, then ds = sqrt(1 + (y')²) dx; if it's x as a function of y, then ds = sqrt(1 + (x')²) dy.

a)

Since the x-axis is the axis of revolution, r is a vertical line segment has length (y - 0) [the zero comes from the fact that the x-axis has equation y=0; if the axis were, for example, the line y=-1, then r would be y-(-1)]. Because y is given as a function of x, let's try x-integration (though we could solve for x as a function of y and do y-integration). Then

r = y(x) = cos(3x) and

ds = sqrt(1 + (y')²) dx = sqrt(1 + 9 sin²(3x)) dx

So the area A is given by

integral from 0 to π/6 of 2π(cos(3x)) sqrt(1 + 9 sin²(3x)) dx

b) is done similarly, except that since the curve is given as x as a function of y, you should try doing this as a y integral. As in (a), r is given by y, and ds = sqrt(1 + (x')²) dy. The area is given by
integral from 1 to 6 of 2πy sqrt(1 + (x')²) dy

where of course you've calculated x' and put it in the integrand.

For c), the curve is given as x in terms of y, so try y-integration. The segment r is horizontal, so r is given by (x-0) [the zero coming from the fact that the axis of revolution is the y-axis, whose equation is x=0]. Because we're doing y-integration, we have
x = x(y) = sqrt(2y - y²), and
ds = sqrt(1 + (x')²) dy. So

A = integral from 10/16 to 17/16 of 2π sqrt(2y - y²) sqrt(1 + (x')²) dy