## MATH1003 Quizzes

Quiz 3: Areas, Volumes and Integration by Substitution
Question 1 Questions
What is the area bounded by the graphs of $f\left(x\right)=sinx$ and $g\left(x\right)=cosx$ and the lines $x=0$ and $x=\frac{\pi }{2}$ ? Exactly one option must be correct)
 a) $2\left(\sqrt{2}-1\right)$ b) $2$ c) $2\sqrt{2}$ d) $1+2\sqrt{2}$ e) None of the above

Choice (a) is correct!
The area is given by $2{\int }_{0}^{\pi ∕4}\phantom{\rule{1em}{0ex}}\left(cosx-sinx\right)\phantom{\rule{1em}{0ex}}dx$ and this equals $2\left(\sqrt{2}-1\right).$
Choice (b) is incorrect
Have you sketched the graphs and found their point of intersection? Try to use symmetry to make your work easier.
Choice (c) is incorrect
Have you sketched the graphs and found their point of intersection? Try to use symmetry to make your work easier.
Choice (d) is incorrect
Have you sketched the graphs and found their point of intersection? Try to use symmetry to make your work easier.
Choice (e) is incorrect
What is the area bounded by the graphs of $f\left(x\right)=\sqrt{x-1}$, $g\left(x\right)=\frac{2}{x}$, the $x$- axis and the line $x=3$ ? Enter your answer correct to 2 decimal places. Do not enter any units.

Correct!
Well done. The required area is ${\int }_{1}^{2}\phantom{\rule{1em}{0ex}}\sqrt{x-1}\phantom{\rule{1em}{0ex}}dx+{\int }_{2}^{3}\phantom{\rule{1em}{0ex}}\frac{2}{x}\phantom{\rule{1em}{0ex}}dx$ and this equals $\frac{2}{3}+2ln\frac{3}{2}\approx 1.48$.
Sketch the curves and find their point of intersection. Now express the required area as the sum of two separate integrals.
Which option below is an integral which equals the volume of the solid of revolution formed when the area between the curves $y=\sqrt{4x}$ and $y=2{x}^{3}$ and between the lines $x=0$ and $x=1$ is rotated about the line $x=3$ ? Exactly one option must be correct)
 a) ${\int }_{0}^{2}\phantom{\rule{1em}{0ex}}2\pi x\left(\sqrt{4x}-2{x}^{3}\right)\phantom{\rule{1em}{0ex}}dx$ b) ${\int }_{0}^{1}\phantom{\rule{1em}{0ex}}4\pi \left(3-x\right)\left(\sqrt{x}-{x}^{3}\right)\phantom{\rule{1em}{0ex}}dx$ c) ${\int }_{0}^{2}\phantom{\rule{1em}{0ex}}2\pi \left(3-x\right)\left(\sqrt{x}-{x}^{3}\right)\phantom{\rule{1em}{0ex}}dx$ d) ${\int }_{0}^{1}\phantom{\rule{1em}{0ex}}2\pi \left(3-x\right)\left({x}^{3}-\sqrt{x}\right)\phantom{\rule{1em}{0ex}}dx$

Choice (a) is incorrect
Which method are you thinking about - shells or discs? Check that you have the right expression to integrate and the right endpoints on the integral sign.
Choice (b) is correct!
This is the integral which gives the required volume when the method of shells is used. The radius of the shell is $3-x$ and the height $\sqrt{4x}-2{x}^{3}$. Values of $x$ range from 0 to 1.
Choice (c) is incorrect
Check the limits on the integral sign.
Choice (d) is incorrect
Which is greater on the interval concerned, $\sqrt{4x}$ or $2{x}^{3}$ ?
The method of shells is used to obtain the volume $V$ of the solid of revolution formed when the area between the curve $y={x}^{2}$ and the $x$-axis, from $x=0$ to $x=1$, is rotated about the line $y=-2$. This method gives us $V=2\pi {\int }_{0}^{1}\phantom{\rule{1em}{0ex}}\left(2+y\right)\left(1-\sqrt{y}\right)\phantom{\rule{1em}{0ex}}dy.$ Which of the integrals below is the one which calculates the same volume by the method of discs? Exactly one option must be correct)
 a) ${\int }_{0}^{1}\phantom{\rule{1em}{0ex}}\pi \left({\left(y+2\right)}^{2}-2\right)\phantom{\rule{1em}{0ex}}dy$ b) ${\int }_{0}^{1}\phantom{\rule{1em}{0ex}}\pi \left(\left({x}^{2}+2\right)-4\right)\phantom{\rule{1em}{0ex}}dx$ c) ${\int }_{0}^{1}\phantom{\rule{1em}{0ex}}\pi \left({\left({x}^{2}+2\right)}^{2}-4\right)\phantom{\rule{1em}{0ex}}dx$ d) ${\int }_{0}^{1}\phantom{\rule{1em}{0ex}}\pi \left({\left({x}^{2}+2\right)}^{2}+4\right)\phantom{\rule{1em}{0ex}}dx$ e) ${\int }_{0}^{1}\phantom{\rule{1em}{0ex}}\pi \left(\left({x}^{2}+2\right)-1\right)\phantom{\rule{1em}{0ex}}dx$

Choice (a) is incorrect
Choice (b) is incorrect
Choice (c) is correct!
Choice (d) is incorrect
Choice (e) is incorrect
Use a substitution to find $\int \phantom{\rule{1em}{0ex}}\frac{t}{\sqrt{4-{t}^{4}}}\phantom{\rule{1em}{0ex}}dt.$ Exactly one option must be correct)
 a) $2\phantom{\rule{1em}{0ex}}sinh\left(\frac{{t}^{2}}{2}\right)+C$ b) $\frac{1}{2}\phantom{\rule{1em}{0ex}}sin\left(2-{t}^{2}\right)+C$ c) $-\frac{1}{2}\phantom{\rule{1em}{0ex}}ln\left(2-{t}^{2}\right)+C$ d) ${cos}^{-1}\left(\frac{{t}^{2}}{2}\right)+C$ e) $\frac{1}{2}\phantom{\rule{1em}{0ex}}{sin}^{-1}\left(\frac{{t}^{2}}{2}\right)+C$

Choice (a) is incorrect
You can check if your answer is right by differentiating it. Do this and you will see why this option is not correct.
Choice (b) is incorrect
You can check if your answer is right by differentiating it. Do this and you will see why this option is not correct.
Choice (c) is incorrect
It is not valid to simplify $\sqrt{4-{t}^{4}}$ to $\sqrt{4}-\sqrt{{t}^{4}}$.
Choice (d) is incorrect
You can check if your answer is right by differentiating it. Do this and you will see why this option is not correct.
Choice (e) is correct!
Useful substitutions are $u={t}^{2}$ or $u=\frac{{t}^{2}}{2}$.
Use a substitution to find $\int \phantom{\rule{1em}{0ex}}\sqrt{tanx}\phantom{\rule{1em}{0ex}}{sec}^{4}x\phantom{\rule{1em}{0ex}}dx.$ Hint: try a substitution that gets rid of the square root sign. Exactly one option must be correct)
 a) $\frac{2}{15}{\left(tanx\right)}^{\frac{3}{2}}{sec}^{5}x+C$ b) $\frac{3}{2}{\left(tanx\right)}^{\frac{3}{2}}+\frac{2}{9}{\left(tanx\right)}^{\frac{9}{2}}+C$ c) $\frac{2}{3}\phantom{\rule{1em}{0ex}}{\left(tanx\right)}^{\frac{3}{2}}+\frac{2}{7}{\left(tanx\right)}^{\frac{7}{2}}+C$ d) $2{sec}^{6}x\sqrt{tanx}+C$ e) None of the above

Choice (a) is incorrect
Differentiate your answer to see why this is incorrect. Try the substitution $u=\sqrt{tanx}$.
Choice (b) is incorrect
Differentiate your answer to see why this is incorrect. Try the substitution $u=\sqrt{tanx}$.
Choice (c) is correct!
The substitution $u=\sqrt{tanx}$, together with the identity $1+{tan}^{2}x={sec}^{2}x$ are useful in this problem.
Choice (d) is incorrect
Differentiate your answer to see why this is incorrect. Try the substitution $u=\sqrt{tanx}$.
Choice (e) is incorrect
Use the shell method to find the volume of the solid formed when the area enclosed by the curves $y={x}^{2}-4x+3$ and $y=-{x}^{2}+2x+3$ is rotated about the $y$-axis. Enter your answer correct to two decimal places. (Do not enter any units.)

Correct!
Well done! Your integral should be $2\pi {\int }_{0}^{3}\phantom{\rule{1em}{0ex}}x\left(-2{x}^{2}+6x\right)\phantom{\rule{1em}{0ex}}dx=27\pi ,$ which is 84.82 correct to two decimal places.
To use the shell method, you need the radius (in this case, just $x$) and the height of the shell (in this case, $\left(-{x}^{2}+2x+3\right)-\left({x}^{2}-4x+3\right)$). Now put this information together to obtain the volume of the shell, and then integrate.
Use the disc method to find the volume of the solid formed when the area enclosed by the curve$y=sinx$ and the $x$-axis, between $x=0$ and $x=\pi$, is rotated about the line $y=2$. Enter your answer correct to two decimal places. (Do not enter any units.)

Correct!
The disc method requires us to set up an integral describing a disc with a hole, formed by rotating the area about the line $y=2$. The outer radius of the disc is 2 and the inner radius is $2-sinx$.
Find the volume of the solid formed when the area enclosed by the curves$y=e{x}^{2}$ and $y={e}^{x}$, and the line $x=0$, is rotated about the $y$-axis. You may use the fact that $\int \phantom{\rule{1em}{0ex}}x{e}^{x}\phantom{\rule{1em}{0ex}}dx={e}^{x}\left(x-1\right)+C.$ Exactly one option must be correct)
 a) $2\pi \left(1-\frac{e}{4}\right)$ b) $2\pi \frac{{e}^{2}}{4}$ c) $2\pi \left(\frac{2e}{3}-1\right)$ d) $\frac{\pi e}{4}$ e) None of the above

Choice (a) is correct!
The volume is given by the integral $2\pi {\int }_{0}^{1}\phantom{\rule{1em}{0ex}}x\left({e}^{x}-e{x}^{2}\right)\phantom{\rule{1em}{0ex}}dx.$ When evaluated using the hint in the question, this equals $2\pi \left(1-\frac{e}{4}$.
Choice (b) is incorrect
Choice (c) is incorrect
Choice (d) is incorrect
Choice (e) is incorrect
What is the volume of the solid obtained when the area between the $x$-axis and the curve $y={x}^{3}+1$, from $x=-1$ to $x=1$, is rotated about the line $y=-1$ ? Exactly one option must be correct)
 a) $\frac{20\pi }{3}$ b) $\frac{10\pi }{3}$ c) $\frac{21\pi }{4}$ d) $4\pi -1$ e) None of the above

Choice (a) is incorrect
Hint: the disc method is easier here than the shell method. The inner radius is 1. What is the outer radius?
Choice (b) is incorrect
Hint: the disc method is easier here than the shell method. The inner radius is 1. What is the outer radius?
Choice (c) is incorrect
Hint: the disc method is easier here than the shell method. The inner radius is 1. What is the outer radius?
Choice (d) is incorrect
Hint: the disc method is easier here than the shell method. The inner radius is 1. What is the outer radius?
Choice (e) is correct!
The correct answer is $\pi {\int }_{-1}^{1}\left({\left({x}^{3}+2\right)}^{2}-1\right)\phantom{\rule{1em}{0ex}}dx=\frac{44\pi }{7}.$ This integral is set up using the disc method. Using the shell method, the integral is $2\pi {\int }_{0}^{2}\left(y+1\right)\left(1-{\left(y-1\right)}^{\frac{1}{3}}\right)\phantom{\rule{1em}{0ex}}dy$, a more difficult integral to evaluate.