##### calculu word problems.

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1.Find the are of the region bounded by the curves y=sin^(-1) [x/4], y=0, and x =4 obtained by integrating with respect to y. Your work must include the definite integral and the anti derivative. 2.set up, but do not evaluate , the integral which gives the volume when the region bounded by the curves y=Ln(x), y=1, ans x=1 is revolved around the line y=-3. 3.don't evalate,the integral which gives the volume when the region bounded by the curves y=Ln(x).y=1and x-=1 is revoled around the line y=-3.

Apr 23rd, 2015

1. The region can be described by the inequalities: 0 <= y <= pi/2 and 0 <= x <= 4 sin y, because the equation y = sin^{-1} [x/4] implies that x/4 = siny,  x = 4 siny and also y = sin^{-1} [4/4] = sin^{-1} [1] = pi/2.

Thus, the area equals the integral ∫0π/2 4sinydy = -4cosy ] 0π/2 = -4cos(π/2) + 4 = 4

We used the antiderivative ∫ 4sinydy = -4cosy + C to evaluate the definite integral.

2. Note that the curves y = ln x and y = 1 intersect at y = e. The region bounded by the given lines is described by inequalities: 1 ≤ x ≤ e, y1(x) = ln x ≤ y ≤ y2(x) = 1. The vertical distance between the graph y = ln x and the line y = -3 equals ln x + 3 and between the graph y = 1 and the line y = -3 is 4.

The volume of the solid obtained by revolving the region around the axis y = -3 equals

π∫1e [(y2(x) + 3)2 – (y1(x) + 3)2]dx = π∫1e [(ln x + 3)2 – 42]dx

Apr 23rd, 2015

The previous solutions need to be corrected. I apologize for the mistakes.

1. The region can be described by inequalities 0 ≤ y ≤ π/2, x1(y) = 4 siny ≤ x ≤ x2(y) = 4

Its area can be expressed by the integral

0π/2 (4 – 4siny)dy = (4y + 4cosy) ] 0π/2 = 4·(π/2) + 4cos(π/2) – 4 = 2π – 4.

We used the antiderivative ∫ (4 – 4siny)dy = 4y + 4cosy + C to evaluate the definite integral.

2. Note that the curves y = ln x and y = 1 intersect at y = e. The region bounded by the given lines is described by inequalities: 1 ≤ x ≤ e, y1(x) = ln x ≤ y ≤ y2(x) = 1. The vertical distance between the graph y = ln x and the line y = -3 equals ln x + 3 and between the graph y = 1 and the line y = -3 is 4.

The volume of the solid obtained by revolving the region around the axis y = -3 equals

π∫1e [(y2(x) + 3)2 – (y1(x) + 3)2]dx = π∫1e [42 – (ln x + 3)2]dx.

The pictures below illustrate both solutions.

Apr 23rd, 2015
04_22_calculus_problems.jpg

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Apr 23rd, 2015
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Apr 23rd, 2015
Dec 4th, 2016
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