W01 Regular

Due date: Sunday 1/18, 11:59pm

Shells

01 $★$

02

Shells volume - set up integrals, both axes

Consider the region in the first quadrant bounded by the lines $x=0$ and $y=2$, and the curve $y=4-x^2$.

Set up integrals to find the volumes of the solids obtained by revolving this region about (i) the $x$-axis, and (ii) the $y$-axis.

(No need to evaluate the integrals in this problem.)

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Solution

Solutions - 0030-02

You can set up each integral using disks/washers or using shells.

(1) Using washers, obtain:

V \;=\; \int_0^\sqrt{2}\pi(4-x^2)^2-\pi\, 2^2\,dx ParseError: Got function '\sqrt' with no arguments as superscript at position 16: V \;=\; \int_0^\̲s̲q̲r̲t̲{2}\pi(4-x^2)^2…

Using shells, obtain:

$$V={\int }_2^{4}2\pi y\sqrt{4-y}dy$$

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(2) Using disks, obtain:

$$V={\int }_2^4\pi {\left(\sqrt{4-y}\right)}^{2}dy$$

Using shells, obtain:

V \;=\; \int_0^\sqrt{2} 2\pi x\left((4-x^2)-2\right)\,dx ParseError: Got function '\sqrt' with no arguments as superscript at position 16: V \;=\; \int_0^\̲s̲q̲r̲t̲{2} 2\pi x\left…Link to original

02 $★$

03

Shells volume - shells v. washers

Consider the region in the $xy$-plane, in the first quadrant, bounded by the $y$-axis on the left, by $y=8-6x^2$ on the top, and $y=2x^2$ on the bottom.

A 3D solid is given by revolving this region around the $y$-axis.

(a) Find the volume of the solid using the method of shells.

(b) Attempt to find the volume of the solid using the method of washers/disks. Why is this harder? (TWO reasons!)

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Solution

Solutions - 0030-03

(a)

(1) Write down shells formula:

$$V={\int }_a^{b}2\pi Rhdr$$

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(2) Define the cross section region:

Bounded above by $y=8-6x^2$. Bounded below by $y=2x^2$.

Bounded left by $x=0$. Bounded right by intersection at line $x=1$.

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(3) Define $R$ and $h$ and $dr$:

$$\begin{array}{cc}R(x)=x,h& =(8-6x^2)-2x^2,dr=dx\\ & \gg \gg h=8-8x^2\end{array}$$

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(4) Plug into shells formula and compute:

$$\begin{array}{cc}V& \gg \gg {\int }_a^{b}2\pi Rhdr\\ & \gg \gg {\int }_0^{1}2\pi (x)(8-8x^2)dx\\ & \gg \gg 16\pi {\int }_0^{1}x-x^{3}dx\\ & \gg \gg 16\pi {[\frac{x^2}{2}-\frac{x^4}{4}]|}_0^1\\ & \gg \gg 16\pi [\frac{1}{2}-\frac{1}{4}]\\ & \gg \gg 4\pi \end{array}$$

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(b)

(1) Write down washers formula using $dy$:

$$V={\int }_a^b\pi (R^2-r^2)dy$$

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(2) Rewrite bounding equations in terms of $y$:

$$\begin{array}{cc}y& =8-6x^2\gg \gg x=\sqrt{\frac{8-y}{6}}\\ & \\ y& =2x^2\gg \gg x=\sqrt{\frac{y}{2}}\end{array}$$

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(3) Determine region boundary data:

Bounded above by $y=8$. Bounded below by $y=0$. Intersection at $y=2$.

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(4) Determine $R$ in two components, with $y=2$ the dividing line:

$$R=\sqrt{\frac{y}{2}}0\le y\le 2$$ $$R=\sqrt{\frac{8-y}{6}}2\le y\le 8$$

Note that $r=0$ for both regions. These are disks.

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(5) Compute the integral:

$$\begin{array}{cc}V& \gg \gg {\int }_a^b\pi (R^2-r^2)dy\\ & \\ & \gg \gg \pi [{\int }_0^2{\left(\sqrt{\frac{y}{2}}\right)}^{2}dy+{\int }_2^8{\left(\sqrt{\frac{8-y}{6}}\right)}^{2}dy]\\ & \\ & \gg \gg \pi [{\int }_0^2\frac{y}{2}dy+{\int }_2^8\frac{8-y}{6}dy]\\ & \\ & \gg \gg \pi [{\left[\frac{y^2}{4}\right]|}_0^2+{[\frac{4}{3}y-\frac{y^2}{12}]|}_2^8]\\ & \\ & \gg \gg \pi [1+[(\frac{32}{3}-\frac{16}{3})-(\frac{8}{3}-\frac{1}{3})]]\\ & \\ & \gg \gg 4\pi \end{array}$$

(6) Why are shells preferable?

1. Only need one integral.
2. Don’t need to rewrite boundary equations in terms of $y$.

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IBP

03 $★$

03

Integration by parts - A and L

Compute the integral:

$$\int x^3\ln xdx$$

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Solution

Solutions - 0010-03

(1) Select $u$ and $v^{\prime }$ considering LIATE:

$$\begin{array}{cc}u=\ln x& v^{\prime }=x^3\\ u^{\prime }={\displaystyle \frac{1}{x}}& v={\displaystyle \frac{x^4}{4}}\end{array}\begin{array}{c}⟶\int u\cdot v^{\prime }=\int x^3\ln x\text{original}\\ ⟶\int u^{\prime }\cdot v=\int {\displaystyle \frac{x^3}{4}}\text{final}\end{array}$$

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(2) Apply IBP formula $uv-\int u^{\prime }\cdot v$:

$$\begin{array}{cc}uv-\int u^{\prime }\cdot v& \gg \gg \frac{x^4}{4}\ln x-\int \frac{x^3}{4}dx\\ & \\ & \gg \gg \frac{x^4}{4}\ln x-\frac{x^4}{16}+C\end{array}$$Link to original

04

04

Integration by parts - A and E

Compute the integral:

$${\int }_0^{3}x{e}^{4x}dx$$

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Solution

Solutions - 0010-04

(1) Select $u$ and $v^{\prime }$ considering LIATE:

$$\begin{array}{cc}u=x& v^{\prime }=e^{4x}\\ u^{\prime }=1& v=\frac{1}{4}{e}^{4x}\end{array}\begin{array}{c}⟶\int u\cdot v^{\prime }=\int x{e}^{4x}\text{original}\\ ⟶\int u^{\prime }\cdot v=\int \frac{1}{4}{e}^{4x}\text{final}\end{array}$$

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(2) Apply IBP formula $uv-\int u^{\prime }\cdot v$ and compute integral:

$$\begin{array}{cc}uv-\int u^{\prime }\cdot v& \gg \gg {\frac{x}{4}{e}^{4x}|}_0^3-{\int }_0^3\frac{1}{4}{e}^{4x}dx\\ & \\ & \gg \gg {\frac{x}{4}{e}^{4x}-\frac{1}{16}{e}^{4x}|}_0^3\\ & \\ & \gg \gg {(\frac{x}{4}-\frac{1}{16})e^{4x}|}_0^3\\ & \\ & \gg \gg e^{12}(\frac{3}{4}-\frac{1}{16})-1(-\frac{1}{16})\\ & \\ & \gg \gg \frac{11}{16}{e}^{12}+\frac{1}{16}\end{array}$$Link to original

05 $★$

05

Integration by parts - A and I

Compute the integral:

$$\int {\tan }^{-1}(x)dx$$

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Solution

Solutions - 0010-05

(1) Select $u$ and $v^{\prime }$ considering LIATE:

$$\begin{array}{cc}u={\tan }^{-1}x& v^{\prime }=1\\ u^{\prime }={\displaystyle \frac{1}{1+x^2}}& v=x\end{array}\begin{array}{c}⟶\int u\cdot v^{\prime }=\int {\tan }^{-1}x\text{original}\\ ⟶\int u^{\prime }\cdot v=\int {\displaystyle \frac{x}{1+x^2}}\text{final}\end{array}$$

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(2) Apply IBP formula $uv-\int u^{\prime }\cdot v$ and compute integral:

$$\begin{array}{cccc}& uv-\int u^{\prime }\cdot v& \gg \gg x{\tan }^{-1}x-\int \frac{x}{1+x^2}dx& \text{(Exp. A)}\end{array}$$

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(3) Perform $u$-sub with $u=1+x^2$ and $du=2xdx$:

$$\begin{array}{cc}\int \frac{x}{1+x^2}dx& \gg \gg \frac{1}{2}\int \frac{2xdx}{1+x^2}\\ & \\ & \gg \gg \frac{1}{2}\int \frac{du}{u}\\ & \\ & \gg \gg \frac{1}{2}\ln |u|+C\\ & \\ & \gg \gg \frac{1}{2}\ln |1+x^2|+C\end{array}$$

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(4) Insert result in Exp. (A):

$$\begin{array}{ccc}& (A)\gg \gg x{\tan }^{-1}x-\frac{1}{2}\ln (1+x^2)+C& \text{(Note B)}\end{array}$$

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Note B: We can change $|1+x^2|$ to $(1+x^2)$ because the inner expression is never negative.

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06

06

Integration by parts - E and T, “breaking the circle”

Compute the integral:

$$\int e^x\sin (x)dx$$

You should perform IBP twice, find an equation, and use algebra to solve it (“breaking the circle”) for the desired integral.

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Solution

Solutions - 0010-06

(1) Select $u$ and $v^{\prime }$ considering LIATE:

$$\begin{array}{cc}u=\sin x& v^{\prime }=e^x\\ u^{\prime }=\cos x& v=e^x\end{array}\begin{array}{c}⟶\int u\cdot v^{\prime }=\int \sin x\cdot e^x\text{original}\\ ⟶\int u^{\prime }\cdot v=\int \cos x\cdot e^x\text{final}\end{array}$$

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(2) Apply IBP formula $uv-\int u^{\prime }\cdot v$ and compute integral:

$$uv-\int u^{\prime }\cdot v\gg \gg \sin x\cdot e^x-\int \cos x\cdot e^{x}dx$$

Therefore:

$$\begin{array}{ccc}& \int \sin x\cdot e^{x}dx=\sin x\cdot e^x-\int \cos x\cdot e^{x}dx& \text{(Eqn. A)}\end{array}$$

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(3) Repeat. Select $u$ and $v^{\prime }$ considering LIATE:

$$\begin{array}{cc}u=\cos x& v^{\prime }=e^x\\ u^{\prime }=-\sin x& v=e^x\end{array}\begin{array}{c}⟶\int u\cdot v^{\prime }=\int \cos x\cdot e^x\text{original}\\ ⟶\int u^{\prime }\cdot v=\int -\sin x\cdot e^x\text{final}\end{array}$$

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(4) Apply IBP formula $uv-\int u^{\prime }\cdot v$ and compute integral:

$$\begin{array}{cc}uv-\int u^{\prime }\cdot v& \gg \gg \cos x\cdot e^x-\int -\sin x\cdot e^{x}dx\\ & \gg \gg \cos x\cdot e^x+\int \sin x\cdot e^{x}dx\end{array}$$

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(5) Now insert in Eqn. A:

$$\int \sin x\cdot e^{x}dx=\sin x\cdot e^x-(\cos x\cdot e^x+\int \sin x\cdot e^{x}dx)$$

Introduce notational label:

$$I=\int \sin x\cdot e^{x}dx$$

Now use this label in Eqn. A and solve for $I$:

$$\begin{array}{c}(A)\gg \gg I=(\sin x-\cos x)e^x-I\\ \\ \gg \gg 2I=(\sin x-\cos x)e^x\\ \\ \gg \gg I=\frac{1}{2}(\sin x-\cos x)e^x+C\end{array}$$Link to original