Integrals for Surface Area, Work, Centroids solved questions

CHAPTER 31
Integrals for Surface Area,
Work, Centroids

SURFACE AREA OF A SOLID OF REVOLUTION
31.1 If the region under a curve y—f(x), above the x- axis, and between x = a and x = b, is revolved about
the *-axis, state a formula for the surface area S of the resulting solid.



[For revolution about the y-axis, change the factor y to x in either integrand.]

31.2 Find the surface area of a sphere of radius r.
Revolve the upper semicircle y= about the jc-axis. Since x~ + y2 = r\ 2x+2yy' = Q,
y' = -x/y, (y'Y = x*ly\ 1 + (y') * = 1 + x'ly = (y- + x~)ly' = rly\ Hence, the surface area S=



In Problems 31.3-31.13, find the surface area generated when the given arc is revolved about the given axis.

31.3 y = x2, 0 < x < j; about the Jt-axis.
y' = 2x. Hence, the surface area Let x = i tan 0, dx = | sec2 0 d6.
So
we get (sec 5 0-sec 3 e)de. By the
reduction formula of Problem 29.39, By Problem 29.40,

Thus we get
In In In
31.4 The same arc as in Problem 31.3, but about the y-axis.

Use Since So



31.5 >> = A:3, O s * < l ; about the x-axis.

so

31.6 about the *-axis.




31.7 in the first quadrant; about the x-axis.

So




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31.8 about the AC-axis.

Then



31.9 1<A:<2; about the y-axis.

Use Then

So




31.10 y2 = l2x, 0 < x < 3 ; about the *-axis.

Now, 2vy' = 12, 4y 2 (y') 2 = 144,
Hence,




31.11 y* + 4x = 2\ny, 0 < y < 3 ; about the ;t-axis.

We have

Thus,




31.12 about the y-axis.

Now,

So In In In

31.13 y = lnjc, l < x < 7 , about the y-axis.

Now, Thus, Let x =
2
tan 0, dx = sec 0 dO. Then By Problem 28.40, this is equal
to In In So we arrive at
In In In In In
We used the fact that

31.14 Find the surface area of a right circular cone of height h and radius of base r.
As is shown in Fig. 31-1, the cone is obtained by revolving about the *-axis the region in the first quadrant
under the line Hence,