calculu word problems.
Calculus

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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.
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 ∫_{0}^{π}^{/2} 4sinydy = 4cosy ] _{0}^{π}^{/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, y_{1}(x) = ln x ≤ y ≤ y_{2}(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∫_{1}^{e} [(y_{2}(x) + 3)^{2} – (y_{1}(x) + 3)^{2}]dx = π∫_{1}^{e} [(ln x + 3)^{2} – 4^{2}]dx
The previous solutions need to be corrected. I apologize for the mistakes.
1. The region can be described by inequalities 0 ≤ y ≤ π/2, x_{1}(y) = 4 siny ≤ x ≤ x_{2}(y) = 4
Its area can be expressed by the integral
∫_{0}^{π}^{/2} (4 – 4siny)dy = (4y + 4cosy) ] _{0}^{π}^{/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, y_{1}(x) = ln x ≤ y ≤ y_{2}(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∫_{1}^{e} [(y_{2}(x) + 3)^{2} – (y_{1}(x) + 3)^{2}]dx = π∫_{1}^{e} [4^{2} – (ln x + 3)^{2}]dx.
The pictures below illustrate both solutions.
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