Regions - AP Calculus AB

Card 0 of 2996

Question

Suppose I want to construct a cylindrical container. It costs 5 dollars per square foot to construct the two circular ends and 2 dollars per square foot for the rounded side. If I have a budget of 100 dollars, what's the maximum volume possible for this container?

Answer

Write down equations representing the volume(V) and cost of the cylinder.

$V=\pi r^2h$

$100=5(2\pi r^2)+2(2\pi rh)=10\pi r^2 +4\pi rh$

We want to find the values of and that will maximize the volume. Before taking the derivative of the volume equation, let's eliminate by using the cost equation.

$h=\frac{100-10\pi r^2}{4\pi r}$

Plug this into the volume equation

$V=\pi r^2\left (\frac{100-10\pi r^2}{4\pi r} \right )=r\left (\frac{100-10\pi r^2}{4} \right )$

$V=\frac{100r-10\pi r^3}{4}$

So now is eliminated in the volume equation. Take the derivative with respect to , set it equal to zero, and solve for .

$\frac{dV}{dr}=\frac{100-30\pi r^2}{4}=0$

$r=\sqrt{\frac{10}{3\pi}}$

We can use this value to find

$h=\frac{100-10\pi \left (\sqrt{\frac{10}{3\pi}} \right )^2}{4\pi \sqrt{\frac{10}{3\pi}}}=\frac{100-\frac{100}{3}}{4\sqrt{\frac{10\pi}{3}}}=\sqrt{\frac{3}{10\pi}} \frac{200}{12}$

$h=\frac{50}{3}\sqrt{\frac{3}{10\pi}}$

Now that we found AND that maximizes the volume, we can find that maximum volume

$V=\pi r^2h=\pi\left (\sqrt{\frac{10}{3\pi}} \right )^2 \frac{50}{3}\sqrt{\frac{3}{10\pi}}$

$V=\frac{50}{3}\sqrt{\frac{10}{3\pi}}$

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Question

We have the function $f(x)=\sqrt{x}$ and it is used to form a three dimensional figure by rotating it about the line $y=4$ . Find the volume of that figure from $x=0$ to $x=5$ .

Answer

Imagine a test rectangle with a length of $4-\sqrt{x}$ and it is rotated around $y=4$ to form a cicular disk with area $\pi (4-\sqrt{x})^2$ . The disk has a thickness $\Delta x$ so that its volume is $\pi(4-\sqrt{x})^2 \Delta x$ . To find the total volume of the figure, turn this into an integral.

$Volume=\pi\int_{0}^{5}(4-\sqrt{x})^2 dx$

Perform the integration

$Volume=\pi\int_{0}^{5}(x-8\sqrt{x}+16) dx=\pi\left ( \frac{1}{2}(5)^2-\frac{16}{3}(5)^{\frac{3}{2}}+16(5) \right )$

$Volume=103.269$

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Question

Consider a volume V of height H along some axis, which we will simply call h, such that . We can express the figure as cross-sectional areas A(h) perpendicular to this h-axis. For example for a cone whose base has radius r, we can choose the axis to go through the point of the cone at h = 0 and then the cone is a stack of little circles of radius , so $A(h) = \pi r^2 h^2/H^2$ .

What is in terms of ?

Answer

The volume is approximated by the Riemann sum formed when we stack n layers atop each other with $\Delta h = H/n$ ,

$V \approx \sum_{k=1}^n A(h_k) ~ \Delta h$

In the limit as n gets large, this expression becomes the integral:

$V = \int_0^H A(h) ~dh$

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Question

What is the volume inside the bowl , ? Hint: This is a solid of revolution about the z-axis with a radius of $\sqrt{x^2 + y^2}$ .

Answer

Since we know that the volume element

$dV = A ~ dz = \pi ~ r^2 ~ dz = \pi ~z ~dz$ ,

and integrating gives:

$V = \int_0^1 \pi ~z~ dz = \pi \left[ \frac{z^2}{2} \right ]_0^1 = \pi \left( \frac{1}{2} - 0 \right ) = \frac{\pi}{2}$

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Question

The following function:

Is rotated around the -axis to create a three-dimensional shape. What is the volume of this object within the interval to ?

Answer

Note that for a given value of x in the function:

The value of f(x) is the distance between the corresponding point on the curve and the x-axis. If the curve is rotated around the x-axis to create a three-dimensional object, f(x) can be seen as the radius of a circular cross-section of the object for any value of x.

To define the volume of this object, we can view it as a sum of infinitely thin disks stacked along the x-axis:

$V = \int \pi(f(x)^{2})dx$

Note how this follows the formula for the volume of a cylinder or disk

$( \pi \cdot height \cdot radius ^2 )$ ,

where is the radius and is the height.

Plugging in our function and range, we can rewrite this as:

$V = \int_{0}^{2} \pi(x+x^{2})^{2}dx$

or

$V = \int_{0}^{2}\pi(x^{4} + 2x^{3} + x^{2})$

Integrating this yields

$V = \pi \left(\frac{x^{5}}{5} + \frac{x^{4}}{2} + \frac{x^{3}}{3}\right)$

Using the upper bound of 2 and lower bound of 0 we find,

$V = \pi\left(\frac{2^{5}}{5}+\frac{2^{4}}{2} +\frac{2^{3}}{3}\right) - \pi\left(\frac{0^{5}}{5}+\frac{0^{4}}{2} +\frac{0^{3}}{3}\right) =\pi \frac{256}{15}$ .

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Question

Using the method of cylindrical disks, find the volume of the region of the graph of

revolved around the -axis on the interval .

Answer

The formula for the volume is given as

$V=\pi\int_{a}^{b}r^2 dx$

where .

As such,

$V=\pi\int_{0}^{10}(x^2)^2 dx=\pi\int_{0}^{10}x^4 dx$ .

When taking the integral, we will use the inverse power rule which states,

$\int x^n=\frac{x^{n+1}}{n+1}$ .

Applying this rule we get

$\pi\int_{0}^{10}x^4 dx=\pi\left(\frac{1}{5}x^5\right)\left.\begin{matrix} , \end{matrix}\right|_{0}^{10}$ .

And by the corollary of the first Fundamental Theorem of Calculus,

$\pi\left(\frac{1}{5}x^5\right)\left.\begin{matrix} , \end{matrix}\right|_{0}^{10}=\pi \left(\left[\frac{1}{5}(10)^5\right]-\left[\frac{1}{5}(0)^5\right]\right)=20000\pi$ .

As such, the volume is

$20000\pi$ units cubed.

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Question

Using the method of cylindrical disks, find the volume of the region of the graph of

$f(x)=\sqrt{x}$

revolved around the -axis on the interval .

Answer

The formula for the volume is given as

$V=\pi\int_{a}^{b}r^2 dx$

where .

As such,

$V=\pi\int_{0}^{3}(\sqrt{x})^2 dx$ .

When taking the integral, we will use the inverse power rule which states,

$\int x^n=\frac{x^{n+1}}{n+1}$ .

Applying this rule we get

$V=\pi\int_{0}^{3}(\sqrt{x})^2 dx=\pi\int_{0}^{3}x, dx=\pi\left(\frac{1}{2}x^2\right)\left.\begin{matrix} , \end{matrix}\right|_{0}^{3}$ .

And by the corollary of the first Fundamental Theorem of Calculus,

$\pi\left(\frac{1}{2}x^2\right)\left.\begin{matrix} , \end{matrix}\right|_{0}^{3}= \pi\left(\left[\frac{1}{2}(3)^2\right]-\left[\frac{1}{2}(0)^2\right]\right)$ .

As such, the volume is

$\frac{9}{2}\pi$ units cubed.

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Question

Find the volume of the equation $y=x^2 ; ; ; 0\leq x \leq 2$ revolved about the -axis.

Answer

This problem can be solved using the Disk Method and the equation

$\int_{a}^{b}\pi [y(x)]^2dx ; ; ; a\leq x\leq b$ .

Using our equation to formulate this equation we get the following.

$\ \int_{0}^{2}\ \pi x^4dx \ \$

Applying the power rule of integrals which states

$\int x^n=\frac{x^{n+1}}{n+1}$

we get,

$\ =\pi \frac{x^{4+1}}{4+1}\ \ =\frac{\pi x^5}{5}|_0^2\ \=\frac{32\pi }{5}$ .

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Question

Find the volume of the area between $f(x)=\sqrt{x}$ and .

Answer

Looking at the graph, we see that the curves intersect at the point and $f(x)=\sqrt{x}$ is on top.

Using the general equation

$\int_{a}^{b}\pi [(f(x))^2-(g(x))^2]dx ; ; ; a\leq x\leq b$ where is on top.

This gives us that the volume is

$\int_{0}^{1}\pi [(\sqrt{x})^2-(x^3)^2]dx=\int_{0}^{1}\pi (x-x^6)dx =\frac{5\pi }{14}$

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Question

Find the volume of the region enclosed by and $y=\sqrt{x}$ rotated about the line .

Answer

The volume can be solved by

$dV=\pi(R^2-r^2)$ where and $r=1-\sqrt[3]{y}$ from 0 to 1 where the curves intersect.

Then, we formulate

$V=\int_{0}^{1}\pi[(1-y^2)^2-(1-\sqrt[3]{y})^2]dy=\frac{13\pi }{30}$

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Question

The unit circle is described by $y=\sqrt{1-x^2}$ and is graphed below. Find the volume of the shaded region rotated around the -axis. The shaded region starts at $x=-\sqrt{2}/2$ and ends at $x=\sqrt{2}/2$ .

Answer

The unit circle is described by $y=\sqrt{1-x^2}$ and is graphed below. Find the volume of the shaded region rotated around the -axis. The shaded region starts at $x=-\sqrt{2}/2$ and ends at $x=\sqrt{2}/2$ .

To find the volume of a region rotated around the -axis, we must find:

$V=\pi\int_a^b{f^2(x)}dx$

In our case, this is:

$V=\pi\int_{\sqrt{2}/2}^{\sqrt{2}/2}{(\sqrt{1-x^2})^2}dx$

$V=\pi\int_{\sqrt{2}/2}^{\sqrt{2}/2}{1-x^2} ; dx$

$V =\pi \left[x-\frac{1}{3}x^3\right]_{-\sqrt{2}/2}^{\sqrt{2}/2}$

$V =\pi \left(\sqrt{2}/2-\frac{1}{3}(\sqrt{2}/2)^3\right) - \pi\left(-\sqrt{2}/2-\frac{1}{3}(-\sqrt{2}/2)^3\right)$

$V = \pi\left(\frac{\sqrt{2}}{2}-\frac{1}{3}\cdot \frac{2\sqrt{2}}{8}\right) - \pi\left(-\frac{\sqrt{2}}{2}-\frac{1}{3}\left(-\frac{2\sqrt{2}}{8}\right) \right )\right)$

$V = \pi \left(\frac{\sqrt{2}}{2}-\frac{2\sqrt{2}}{24} \right )-\pi\left(-\frac{\sqrt{2}}{2} +\frac{2\sqrt{2}}{24}\right )$

$V = \pi \left(\frac{12\sqrt{2}}{24}-\frac{2\sqrt{2}}{24} \right )-\pi\left(-\frac{12\sqrt{2}}{24} +\frac{2\sqrt{2}}{24}\right )$

$V = \pi \left(\frac{10\sqrt{2}}{24} \right )-\pi\left(-\frac{10\sqrt{2}}{24} \right )$

$V = \frac{20\pi}{24}\sqrt{2}$

$V = \frac{5\pi}{6}\sqrt{2}$

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Question

Determine the volume for the region bounded by and if the region is revolved about the -axis.

Answer

In the region bounded by and , write the formula to determine the volume of a solid revolved about an axis. This is the washer method.

$V=\int_{a}^{b} A(x):dx= \pi \int_{a}^{b} (R(x)^2-r(x)^2):dx$

Determine which function is the top function. This will be the big radius, and the small radius is represented by the bottom curve. The bounds of the integral are at the intersections of the two bounded functions. Evaluate the integral.

$V=\pi \int_{0}^{1} ((x)^2-(x^2)^2):dx$

$V=\pi \int_{0}^{1} x^2-x^4:dx$

$V=\pi\left[\frac{x^3}{3}-\frac{x^5}{5}\right]_{0}^{1}=\pi\left[\frac{1}{3}-\frac{1}{5}\right] = \frac{2\pi}{15}$

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Question

Find the volume of the solid shape made by rotating about the -axis on the interval .

Answer

Find the volume of the solid shape made by rotating f(x) about the x-axis on the interval

Recall the following formula for volume of a solid shape:

$V=\int_{a}^{b}A(x)dx$

Where A(x) is equal to the the area of a disk made by rotating our function:

$A(x)=\pi r^2$

So we need to put it all together:

$A(x)=\pi(x^4+6)^2=\pi(x^8+12x^4+36)$

Then, we know that our limits of integration must be 0 and 3, because they are the interval that we are working with.

$V=\pi \int_{0}^{3}x^8+12x^4+36dx$

$V=\pi \left(\frac{x^9}{9}+\frac{12x^5}{5}+36x\right) _0 ^3$

(Notice that plugging in 0 will yield 0, so we only need to really worry about the 3)

$V=\pi \left(\frac{3^9}{9}+\frac{12(3)^5}{5}+36(3)\right)=\frac{14391 \pi}{5}$

So our answer is:

$\frac{14391 \pi}{5}$

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Question

Find the volume of the solid generated by rotating the shape bounded between the functions and .

Answer

The first step is to find the lower and upper bounds, the points where the two functions intersect:

Knowing these, the solid generated will have the following volume:

$V=\int_{-3}^{3}(\pi(9)^2-\pi(x^2)^2)dx=\int_{-3}^{3}(\pi(81)-\pi(x^4))dx=\pi\left(81x-\frac{x^5}{5}\right)|_{-3}^3$

$V=\pi\left(81(3)-\frac{243}{5})-(81(-3)+\frac{243}{3}\right)=\frac{1944}{5}\pi$

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Question

What is the volume of the solid formed when the function

$y=\frac{3x^2}{2}$

is revolved around the line over the interval ?

Answer

The disc method can be used to find the volume of this solid of revolution. To use the disc method, we must set up and evaluate an integral of the form

$\pi \int_{a}^{b}[R(x)]^2dx$

where a and b are the endpoints of the interval, and R(x) is the distance between the function and the rotation axis.

First find R(x):

$R(x)=\frac{3x^2}{2}-0=\frac{3x^2}{2}$

Next set up and evaluate the integral:

$\pi \int_{0}^{2}\frac{3x^2}{2}dx=\pi \left(\frac{3(2)^3}{6}-\frac{3(0)^3}{6} \right)$

Simplifying, we find that the volume of the solid is

$\pi \left(\frac{3(8)}{6}-0 \right)=4\pi$

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Question

For any given geometric equation with one variable, volume over a given region is defined as the definite integral of the surface area over that specific region.

Given the equation for surface area of any sphere, $A(r)=4\pi*r^2$ , determine the volume of the piece of the sphere from to .

Answer

Recall that volume is the definite integral of surface area over a given region.

$V(r)=\int_{a}^{b} A(r)dr$

Given our surface area formula, we can determine the volume in one step.

$V(r)=\int_{1}^{5}4\pi r^2 dr= \frac{4}{3}\pi* 5^3- \frac{4}{3}\pi* 1^3= \frac{496\pi}{3}$

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Question

What is the volume of the solid formed by revolving the curve $y=\sqrt{x}$ about the on $x\in(0,4)$ ?

Answer

We take the region and think about revolving rectangles of height $\sqrt{x}$ and width $\Delta x$ about the line . This forms discs of volume $\pi(\sqrt{x})^2\Delta x$ . We then sum all of these discs on the interval and take the limit as the number of discs on the interval approaches infinity to arrive at the definite integral $\pi\int_{0}^{4}(\sqrt{x})^2dx=\pi\int_{0}^{4}xdx=\pi(\frac{x^{2}}{2})|_{0}^{4}=\pi(4^{2}/2)=8\pi$

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Question

Find the volume of the solid generated by rotating the curve , for the range around the x-axis.

Answer

To solve this problem, treat the function

As the radius of a disc with infinitesimal thickness . This disc would then have a volume defined the surface area:

$\pi r^2:\pi(3x^2+2x)^2$

$\pi(9x^4+12x^3+4x^2)$

And a thickness or height:

The volume of the solid generated would be the sum of these discs, i.e. the integral:

$V=\pi \int_{0}^{2}(9x^4+12x^3+4x^2)dx=\pi(\frac{9x^5}{5}+3x^4+\frac{4x^3}{3})|_0^2$

$V=\pi(\frac{288}{5}+48+\frac{32}{3})-(0+0+0)$

$V=\frac{1744\pi}{15}$

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Question

Find the volume generated by rotating the region between and around the -axis.

Answer

Since the functions are being rotated around the y-axis, they should be rewritten. It may help to write them in terms of y:

$h(y)=\sqrt{y}$

And

$j(y)=\frac{y}{2}$

Here, function designations have been used solely to eliminate the ambiguity of using x for both functions.

Now, there are two points of intersection for these functions:

$\sqrt y= \frac{y}{2}:y=0,4$

These will be the bounds for the integration to follow.

Finally, treat these two functions as the radii for two regions: a disk from the larger radius, and a hole from the smaller radius, allowing for the formation of a ring with a cross-sectional area:

$A=\pi(\sqrt y^2 - \frac{y^2}{4})$

These disks have infinitesimal thickness , and so the volume of the solid can be found by adding all of these disks together; i.e. by taking the integral:

$V=\int Ady=\int_{0}^{4}\pi(y-\frac{y^2}{4})dy$

$V=\pi(\frac{6y^2-y^3}{12})|_0^4$

$V=\frac{8}{3}\pi$

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Question

Find the volume generated by rotating the function $f(x)=3xe^{x^3}$ around the -axis over the interval .

Answer

To approach this problem, treat the function $f(x)=3xe^{x^3}$ as the radius of a disc, with infintesimal thickness dx (which has the x-axis going through its center).

This infinitesismal volume of this disc is:

$dV=\pi(3xe^{x^3})^2dx$

$dV=9\pi x^2e^{2x^3}dx$

Now the volume of the solid rotated around the x-axis is the sum of these discs. In other word, it is the integral:

$V=9\pi \int_0^2 x^2e^{2x^3}dx$

Note that the derivative of $e^{2x^3}$ is $6x^2e^{2x^3}$ , so from this we can infer that the integral of $x^2e^{2x^3}$ is $\frac{1}{6}e^{2x^3}$ :

$V=\frac{9}{6}\pi(e^{2x^3})|_0^2$

$V=\frac{9}{6}\pi(e^{16}-e^0)$

$V=\frac{3}{2}\pi(e^{16}-1)$

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