Consider the function f(x) dened on the closed interval [-2; 6 ] given by: g(x)
ID: 2846175 • Letter: C
Question
Consider the function f(x) dened on the closed interval [-2; 6 ] given by:
g(x) =(12 + 4x -x^2)^1/2
1. Rewrite the function in the form
(b^2-(x - a)^2)^1/2
2. Use an appropriate trigonometric substitution to evaluate the integral:
Hint: Use your result from part (1) and clearly show your work for full credit.
Integrals of upper bound 6 and lower bound -2 of ((12 + 4x -x^2)^1/2)dx
3. Graph the function on the grid below.
4. What shape is graphed? Compute the area described by the integral directly and compare your result to that of the
trigonometric substitution in 2.
5. If the region between g(x) and y = 0 on closed interval [??2; 6 ] is rotated about the x-axis, what solid is formed?
6. Compute the volume of the solid in the question above by:
i. the disk method ii. a formula from geometry
Explanation / Answer
Completing the squares, we obtain
[(12 + (4 - 4) + 4x - x^2)^{1/2} = (16 - (4 - 4x + x^2))^{1/2} = (16 - (x - 2)^2)^{1/2}]
which is equivalent to
[(16 - (2 - x)^2)^{1/2}]
We want to evaluate
[int_{-2}^6 (16 - (x - 2)^2)^{1/2},dx]
Let (u = x - 2). Then, (du = dx). This yields:
[int_{-2}^6 (16 - u^2)^{1/2},du]
By trigonometric substitution, we let (4sin( heta) = sqrt{16 - u^2}), which comes from (sin( heta) = dfrac{sqrt{16 - u^2}}{4}). Also let (cos( heta) = dfrac{u}{4}), which is equivalent to (4cos( heta) = u). Then, (du = -4sin( heta),d heta). This gives:
[int_{-2}^6 (4sin( heta))cdot (-4sin( heta),d heta)]
which is simplified as
[int_{-2}^6 -16sin^2( heta),d heta = -16 int_{-2}^6 sin^2( heta),d heta]
Recall that a trig identity is
[cos(2 heta) = 1 - 2sin^2( heta)]
Solve for (sin^2( heta)) to get
[sin^2( heta) = dfrac{1}{2}(1 - cos(2 heta))]
So we have
[-16 int_{-2}^6 dfrac{1}{2}(1 - cos(2 heta)),d heta = -8 int_{-2}^6 (1 - cos(2 heta)),d heta]
Therefore, we obtain
[left.-8left( heta - dfrac{1}{2}sin(2 heta) ight) ight ert_{x = -2}^{x = 6}]
Since
[ heta = rcsinleft(dfrac{sqrt{16 - u^2}}{4} ight) = rcsinleft(dfrac{sqrt{16 - (x - 2)^2}}{4} ight)]
and
[sin(2 heta) = 2sin( heta)cos( heta) = 2dfrac{sqrt{16 - u^2}}{4}dfrac{u}{4} = dfrac{(x - 2)sqrt{16 - (x - 2)^2}}{8}]
we obtain
[left.-8left( rcsinleft(dfrac{sqrt{16 - (x - 2)^2}}{4} ight) + dfrac{(x - 2)sqrt{16 - (x - 2)^2}}{8} ight) ight ert_{x = -2}^{x = 6}]
Thus, by the Fundamental Theorem of Calculus, we have (8pi) [This is left for you as an exercise].
Here is the link to the graph shaded at the given interval: http://www.wolframalpha.com/input/?i=integrate+%2816+-+%28x+-+2%29%5E2%29%5E%281%2F2%29+from+-2+to+6
The part of the circle is graphed based on Wolfram.
Notice that the radius of that circle is (4). Since the shape is precisely half-circle,
[A = dfrac{1}{2}pi r^2]
Thus,
[A = dfrac{1}{2}pi (4)^2 = 8pi]
We know that the graph of the shaded region is the half-circle. Revolve it about the (x)-axis, so we obtain the sphere.
It's a lot easier to compute the volume by geometry rather than disk method. Here is the approach:
[V = dfrac{4}{3}pi r^3 = dfrac{4}{3}pi (4)^3 = dfrac{256pi}{3}]
By disk method, we have...
[V = int_{-2}^6 pi(sqrt{16 - (x - 2)^2})^2,dx = int_{-2}^6 pi(16 - (x - 2)^2),dx]
Let (u = x - 2). Then, (du = dx). So...
[V = pi int_{-2}^6 (16 - u^2),du]
[V = left.pileft(16u - dfrac{1}{3}u^3 ight) ight ert_{x = -2}^{x = 6}]
[V = left.pileft(16(x - 2) - dfrac{1}{3}(x - 2)^3 ight) ight ert_{x = -2}^{x = 6}]
So by the Fundamental Theorem of Calculus, we obtain the equivalent value (dfrac{256pi}{3}).