##### calculu word problems.

*label*Calculus

*account_circle*Unassigned

*schedule*1 Day

*account_balance_wallet*$5

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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