# Thread: Calculus Old Quiz Problems Help

1. ## Calculus Old Quiz Problems Help

1. Let R be the region bounded by the graphs of y= sinx and y = 1 - sinx on (pi/6, 5pi/6)
Use the washer or disk method to set up (but do not evaluate) an integral that will determine the volume of the solid obtained by revolving R about the line y=3

2. Use calculus determine volume of the solid generated when R is revolved around the x-axis?

Let R be the region bounded by x=y^2 -4y + 4 and the vertical lines x=0 and x=1.

2. ## Re: Calculus Old Quiz Problems Help

1. Use the washer method. The radius of the smaller circle will be:

$3-\sin x$

The radius of the larger circle will be:

$3-(1-\sin x)=2+\sin x$

2. If you want to use the washer method, you will need to solve for $y$ in terms of $x$. You can also use the cylindrical shell method.

To use the washer method:

$$y^2-4y+4-x=0$$

$$y = \dfrac{4 \pm \sqrt{4^2-4(4-x)}}{2} = 2\pm \sqrt{x}$$

So, our larger radius is $2+\sqrt{x}$ and our smaller radius is $2-\sqrt{x}$. This gives:

$$\pi \int_0^1 \big[ (2+\sqrt{x})^2 - (2-\sqrt{x})^2\big]dx = \pi \int_0^1 8\sqrt{x}dx = \dfrac{16\pi}{3}$$

To use the cylindrical shell method, we have $h=y$ and $r=1-x = 1-(y^2-4y+4) = -y^2+4y-3$. Additionally, we need lower and upper bounds for $y$. So, we need to solve

$$0 = y^2-4y+4$$

and

$$1=y^2-4y+4$$

which gives $y=2$ or $y=1,y=3$. We use the latter choice (because it gives both a lower and upper bound).

So, we have:

$$\int_1^3 2\pi y (-y^2+4y-3)dy = 2\pi \int_1^3 (-y^3+4y^2-3y)dy = 2\pi\left( -\dfrac{80}{4}+\dfrac{4\cdot 26}{3} - \dfrac{3\cdot 8}{2} \right) = \dfrac{16\pi}{3}$$

Note: Graphing this is useful to help determine the region that is bounded. Without graphing, I would not have known that the radius of each shell was $1-x$.

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