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Calculus 1 : Calculus

Study concepts, example questions & explanations for Calculus 1

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Example Questions

Example Question #3931 : Calculus

Find the volume generated by rotating the region between f(x)=x^2 and g(x)=2x around the -axis.

Possible Answers:

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

$\frac{8}{15}\pi$

$\frac{8}{15}$

$\frac{4}{3}\pi$

$\frac{8}{3}$

Correct answer:

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

Explanation:

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

y=x^2

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

And

y=2x

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

Here, function designations [h(y),j(y)] 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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Example Question #3931 : Calculus

Find the volume of the function

f(x)=5

revolved around the -axis on the interval [0, 10] .

Possible Answers:

$125\pi$ cubic units

$210\pi$ cubic units

$500\pi$ cubic units

$250\pi$ cubic units

Correct answer:

$250\pi$ cubic units

Explanation:

The formula for the volume of the region revolved around the x-axis on the interval [a,b] is given as

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

where r=f(x) .

As such,

$V=\pi \, \int_{0}^{10} 5^2 \, dx$ .

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

$\pi \, \int_{0}^{10} 5^2 \, dx=\pi \, \int_{0}^{10} 25 \, dx$

$=\pi (25x)|_{0}^{10}$ .

And by the corollary of the First Fundamental Theorem of Calculus

$\pi (25x)|_{0}^{10}=\pi[25(10)-25(0)]=250\pi$ .

As such, the volume is

$250\pi$ cubic units.

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Example Question #3932 : Calculus

Find the volume of the function $f(x)=\sqrt{cos \,x}$ revolved around the -axis on the interval $\left[0, \frac{\pi}{2}\right]$ .

Possible Answers:

$\pi^2$ cubic units

$\pi$ cubic units

$\frac{\pi}{2}$ cubic units

$3.14\pi$ cubic units

Correct answer:

$\pi$ cubic units

Explanation:

The formula for the volume of the region revolved around the x-axis on the interval [a,b] is given as

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

where r=f(x) .

As such,

$V=\pi \, \int_{0}^{\pi/2} (\sqrt{cos\,x})^2 \, dx$

$=\pi \, \int_{0}^{\pi/2} cos\,x \, dx$

$=\pi(sin\,x)|_{0}^{\pi/2}$ .

And by the corollary of the First Fundamental Theorem of Calculus

$\pi(sin\,x)|_{0}^{\pi/2}=\pi[sin(\frac{\pi}{2})-sin(0)]=\pi$ .

As such, the volume is

$\pi$ cubic units.

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Example Question #21 : Regions

Find the volume generated by rotating the function $f(x)=3xe^{x^3}$ around the -axis over the interval x=[0,2] .

Possible Answers:

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

$\frac{3}{2}\pi(e^{8}-1)$

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

$\frac{3}{2}\pi(e^{16})$

$\frac{3}{2}\pi(e^{8})$

Correct answer:

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

Explanation:

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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Example Question #3931 : Calculus

Find the volume of the solid generated by rotating the parabola f(x)=25-x^2 around the -axis, over the region where the function is greater than or equal to zero.

Possible Answers:

$\frac{5000}{3}\pi$

$\frac{10000}{3}\pi$

$5000\pi$

$10000\pi$

Correct answer:

$\frac{10000}{3}\pi$

Explanation:

In stating the bounds, the question is asking that we use those values of x where the function intercepts the x-axis:

25-x^2=0

$x=\pm 5$

Now, treat the function f(x)=25-x^2 as the radius of a disc with infinitesimal thickness . The volume of this disc is thus

dV=Adx

$dV=\pi f(x)^2 dx$

$dV=\pi (x^4-50x^2+625)dx$

And the volume can found via integration:

$V= \int_{-5}^5\pi(x^4-50x^2+625)dx$

$V=\pi (\frac{x^5}{5}-\frac{50x^3}{3}+625x)|_{-5}^5$

$V=\pi (\frac{3125}{5}-\frac{6250}{3}+3125)-\pi(-\frac{3125}{5}+\frac{6250}{3}-3125)$

$V=\frac{10000}{3}\pi$

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Example Question #3934 : Calculus

For this problem assume that volume is given by:

$V=\int S(x)dz$ , where S(x) is the surface area function.

Given the formula for surface area of a structure is given by S(x)=32sec^2(x) .

Determine the volume function of this structure.

For this problem assume that is not a function of . given below is a constant.

Possible Answers:

V=32sec^2(x)*z +C

V=32tan(x)(z+C)

V=32sec^2(x)(z+C)

V=z(32tan(x)+C)

Correct answer:

V=32sec^2(x)(z+C)

Explanation:

We're given that

$V=\int S(x)dz$

Since the indefinite integral is with respect to , we can move the surface area function in front of the integral since we are told that is not a function of .

$V=S(x)\int 1dz$

Using the power rule,

$\int 1dz=z+C$ , where is a constant.

V=S(x)(z+C)

In our problem S(x)=32sec^2(x)

Therefore, volume can be given as

V=S(x)(z+C)=32sec^2(x)(z+C)

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Example Question #2908 : Functions

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

f(x)=x-5

on the interval [5,10] .

Possible Answers:

$150\pi$ units cubed

$250\pi$ units cubed

$300\pi$ units cubed

$\frac{125}{3}\pi$ units cubed

Correct answer:

$\frac{125}{3}\pi$ units cubed

Explanation:

The formula for volume is given as

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

where r=f(x) .

As such,

$V=\pi \int_{5}^{10} (x-5)^2\,dx =\pi \int_{5}^{10}x^2-10x+25\,dx$ .

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 term by term we get

$\pi \int_{5}^{10}x^2-10x+25\,dx =\pi \left ( \frac{x^3}{3} - \frac{10x^2}{2} + 25x\right) |_{5}^{10}$

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

$\pi \left ( \frac{x^3}{3} - \frac{10x^2}{2} + 25x\right |_{5}^{10}=\pi \left ( [\frac{10^3}{3}-\frac{10(10)^2}{2}+25(10)]-[\frac{5^3}{3}-\frac{10(5)^2}{2}+25(5)]\right )$

$=\frac{125}{3}\pi$

As such, the volume is

$\frac{125}{3}\pi$ units cubed.

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Example Question #2909 : Functions

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

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

on the interval $[\pi,9+\pi]$ .

Possible Answers:

$50\pi$ units cubed

$55\pi$ units cubed

$35\pi$ units cubed

$40.5}\pi$ units cubed

Correct answer:

$40.5}\pi$ units cubed

Explanation:

The formula for volume is given as

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

where r=f(x) .

As such,

$V=\pi \int_{\pi}^{9+\pi} (\sqrt{x-\pi})^2\,dx =\pi \int_{\pi}^{9+\pi} x-\pi\,dx$ .

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 term by term we get

$\pi \int_{\pi}^{9+\pi} x-\pi\,dx=\pi \left ( \frac{x^2}{2}-\pi x\ \right) |_{\pi}^{9+\pi}$ .

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

$\pi \left ( \frac{x^2}{2}-\pi x\ \right )|_{\pi}^{9+\pi}=\pi\left ( [\frac{(9+\pi)^2}{2}-\pi (9+\pi)]-[\frac{\pi^2}{2}-\pi^2] \right )$

$=40.5\pi$ .

As such, the volume is

$40.5}\pi$ units cubed.

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Example Question #2910 : Functions

Set up the integral to find the volume of the solid if the region is bounded by y=x^2 and y=4x-3 , revolved about x=5 .

Possible Answers:

$\int_{1}^{3} 2\pi(5-x)(4x-3-x^2)\:dx$

$\int_{1}^{5} \pi[(4x-3)^2-x^2]^2 \:dx$

$\int_{1}^{3} 2\pi(x-5)(4x-3-x^2)\:dx$

$\int_{1}^{5} \pi(4x-3-x^2)^2\:dx$

$\int_{1}^{3} \pi(5-x)^2(4x-3-x^2)\:dx$

Correct answer:

$\int_{1}^{3} 2\pi(5-x)(4x-3-x^2)\:dx$

Explanation:

For this problem, we will use the shell method. Write the formula for the shell method.

$V=\int_{a}^{b} 2\pi rh \:dr$

The thickness of the shell is .

The radius of the shell is the distance from the y-axis to the vertical rectangles, 5-x .

The height of the shell is the top curve subtract the bottom curve. The top curve is y=4x-3 and the bottom curve is y=x^2 . The height is: 4x-3-x^2 .

Determine the bounds by setting the top and bottom equations equal to each other.

x^2=4x-3

Solve for roots.

x^2-4x+3=0

(x-3)(x-1)=0

The roots are x = 1,3 and will be the bounds to the integral. Write the integral.

$V=\int_{a}^{b} 2\pi rh \:dr= \int_{1}^{3} 2\pi(5-x)(4x-3-x^2)\:dx$

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Example Question #3935 : Calculus

Sphere segment

A sphere with a radius of 5 has a segment with perpendicular planes cut out of it, such as is pictured above.

If the first cut is 4 units to the left of the origin and the second cut is 2 units to the left of the origin, what is the volume of the segment?

Possible Answers:

$\frac{49}{3}\pi$

$\frac{128}{3}\pi$

$\frac{94}{3}\pi$

$\frac{251}{3}\pi$

$\frac{23}{3}\pi$

Correct answer:

$\frac{94}{3}\pi$

Explanation:

To approach this problem, begin by picturing how the sphere might be projected onto the two-dimensional Cartesian plane:

Sphere segment outlined

In its two-dimensional projection, we notice the outline of a circle. The formula for a circle of radius in the Cartesian coordinate system is given as:

x^2+y^2=r^2

Which can be rewritten in terms of as $y=\sqrt{r^2-x^2}$

Now note the disk method of volume creation wherein we rotate a function around an axis (for instance, the x-axis):

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

The new function in the integral is akin to the formula of the volume of a cylinder:

$\pi hr_{cylinder}^2$ where $r_{cylinder}^2\rightarrow f^2(x)$ and $h \rightarrow dx$

The integral sums up these thin cylinders to give the volume of the shape.

Treating as our f(x) , this integral can be written as:

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

$V=\pi(r^2x-\frac{x^3}{3})|_{a}^{b}$

Consider the points where circle contacts the x-axis, the greatest and smallest possible values of on the circle:

$V=\pi(r^2x-\frac{x^3}{3})|_{-r}^{r}$

$V=\pi(r^2r-\frac{r^3}{3})-\pi(r^2(-r)-\frac{(-r)^3}{3})$

$V=\pi(\frac{2}{3}r^3)-\pi(-\frac{2}{3}r^3)$

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

This is how the volume of a sphere can be derived, and similary, this is how we can derive the area of our segment:

$V=\pi(5^2x-\frac{x^3}{3})|_{-4}^{-2}$

$V=\pi(5^2(-2)-\frac{(-2)^3}{3})-\pi(5^2(-4)-\frac{(-4)^3}{3})$

$V=\frac{94}{3}\pi$

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Calculus 1 : Calculus

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #3931 : Calculus

Example Question #3931 : Calculus

Example Question #3932 : Calculus

Example Question #21 : Regions

Example Question #3931 : Calculus

Example Question #3934 : Calculus

Example Question #2908 : Functions

Example Question #2909 : Functions

Example Question #2910 : Functions

Example Question #3935 : Calculus

All Calculus 1 Resources

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