6.2 Volumes/15 - Revision history

Cynthiac61995@students.laalliance.org at 17:14, 13 December 2022

2022-12-13T17:14:05Z

← Older revision		Revision as of 17:14, 13 December 2022
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	<math>		<math>
	\begin{align}		\begin{align}
	\pi\int_{-1}^{1}[\left(1-y^2\right)^2~~\right]~~ dy & = \pi\int_{-1}^{1}\left(1-2y^2+y^4\right)dy \\[2ex]		\pi\int_{-1}^{1}\left(1-y^2\right)^{2} dy & = \pi\int_{-1}^{1}\left(1-2y^2+y^4\right)dy \\[2ex]

	&= \pi\left[y-\frac{2}{3}y^3+\frac{y^5}{5}\right]\Bigg\|_{-1}^{1} \\[2ex]		&= \pi\left[y-\frac{2}{3}y^3+\frac{y^5}{5}\right]\Bigg\|_{-1}^{1} \\[2ex]

Cynthiac61995@students.laalliance.org at 17:13, 13 December 2022

2022-12-13T17:13:40Z

← Older revision		Revision as of 17:13, 13 December 2022
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	<math>		<math>
	\begin{align}		\begin{align}
	\pi\int_{-1}^{1}\left(1-y^2\right)^2 dy & = \pi\int_{-1}^{1}\left(1-2y^2+y^4\right)dy \\[2ex]		\pi\int_{-1}^{1}[\left(1-y^2\right)^2\right] dy & = \pi\int_{-1}^{1}\left(1-2y^2+y^4\right)dy \\[2ex]

	&= \pi\left[y-\frac{2}{3}y^3+\frac{y^5}{5}\right]\Bigg\|_{-1}^{1} \\[2ex]		&= \pi\left[y-\frac{2}{3}y^3+\frac{y^5}{5}\right]\Bigg\|_{-1}^{1} \\[2ex]

Cynthiac61995@students.laalliance.org at 17:11, 13 December 2022

2022-12-13T17:11:45Z

← Older revision		Revision as of 17:11, 13 December 2022
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	<math>		<math>
	\begin{align}		\begin{align}
	\pi\int_(-1)^1\left(1-y^2\right)^2 dy & = \pi\int_(-1)^1\left(1-2y^2+y^4\right)dy \\[2ex]		\pi\int_{-1}^{1}\left(1-y^2\right)^2 dy & = \pi\int_{-1}^{1}\left(1-2y^2+y^4\right)dy \\[2ex]

	&= \pi\left[y-\frac{2}{3}y^3+\frac{y^5}{5}\right]\Bigg\|_(-1)^1 \\[2ex]		&= \pi\left[y-\frac{2}{3}y^3+\frac{y^5}{5}\right]\Bigg\|_{-1}^{1} \\[2ex]
	&= \pi\left[1+\frac{2}{3}+\frac{1}{5}\right]- \pi\left[-1+\frac{2}{5}-\frac{1}{5}\right] \\[2ex]		&= \pi\left[1+\frac{2}{3}+\frac{1}{5}\right]- \pi\left[-1+\frac{2}{5}-\frac{1}{5}\right] \\[2ex]
	&= \frac{16\pi}{15}		&= \frac{16\pi}{15}

Cynthiac61995@students.laalliance.org at 17:07, 13 December 2022

2022-12-13T17:07:47Z

← Older revision		Revision as of 17:07, 13 December 2022
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	<math>		<math>
	\begin{align}		\begin{align}
	\pi\int_-1^1\left(1-y^2\right)^2 dy & = \pi\int_-1^1\left(1-2y^2+y^4\right)dy \\[2ex]		\pi\int_(-1)^1\left(1-y^2\right)^2 dy & = \pi\int_(-1)^1\left(1-2y^2+y^4\right)dy \\[2ex]

	&= \pi\left[y-\frac{2}{3}y^3+\frac{y^5}{5}\right]\Bigg\|_-1^1 \\[2ex]		&= \pi\left[y-\frac{2}{3}y^3+\frac{y^5}{5}\right]\Bigg\|_(-1)^1 \\[2ex]
	&= \pi\left[1+\frac{2}{3}+\frac{1}{5}\right]- \pi\left[-1+\frac{2}{5}-\frac{1}{5}\right] \\[2ex]		&= \pi\left[1+\frac{2}{3}+\frac{1}{5}\right]- \pi\left[-1+\frac{2}{5}-\frac{1}{5}\right] \\[2ex]
	&= \frac{16\pi}{15}		&= \frac{16\pi}{15}

Cynthiac61995@students.laalliance.org at 17:06, 13 December 2022

2022-12-13T17:06:44Z

← Older revision		Revision as of 17:06, 13 December 2022
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	<math>		<math>
	\begin{align}		\begin{align}
	\pi\int_-1^1\left(1-y^2\right)^2 dy &= \pi\int_-1^1\left(1-2y^2+y^4\right)dy \\[2ex]		\pi\int_-1^1\left(1-y^2\right)^2 dy & = \pi\int_-1^1\left(1-2y^2+y^4\right)dy \\[2ex]

	&= \pi\left[y-\frac{2}{3}y^3+\frac{y^5}{5}\right]\Bigg\|_-1^1 \\[2ex]		&= \pi\left[y-\frac{2}{3}y^3+\frac{y^5}{5}\right]\Bigg\|_-1^1 \\[2ex]
	&= \pi\left[1+\frac{2}{3}+\frac{1}{5}\right]- \~~pileft~~[-1+\frac{2}{5}-\frac{1}{5}\right] \\[2ex]		&= \pi\left[1+\frac{2}{3}+\frac{1}{5}\right]- \pi\left[-1+\frac{2}{5}-\frac{1}{5}\right] \\[2ex]
	&= \frac{16\pi}{15}		&= \frac{16\pi}{15}

	\end{align}		\end{align}
	</math>		</math>

Cynthiac61995@students.laalliance.org: Created page with " $x=y^2, x=1; \text{about x=1}$ $\begin{align} \pi\int_-1^1\left(1-y^2\right)^2 dy &= \pi\int_-1^1\left(1-2y^2+y^4\right)dy \\[2ex] &= \pi\left[y-\frac{2}{3}y^3+\frac{y^5}{5}\right]\Bigg|_-1^1 \\[2ex] &= \pi\left[1+\frac{2}{3}+\frac{1}{5}\right]- \pileft[-1+\frac{2}{5}-\frac{1}{5}\right] \\[2ex] &= \frac{16\pi}{15} \end{align}$ "

2022-12-13T17:05:29Z

Created page with "<math> x=y^2, x=1; \text{about x=1} </math> <math> \begin{align} \pi\int_-1^1\left(1-y^2\right)^2 dy &= \pi\int_-1^1\left(1-2y^2+y^4\right)dy \\[2ex] &= \pi\left[y-\frac{2}{3}y^3+\frac{y^5}{5}\right]\Bigg|_-1^1 \\[2ex] &= \pi\left[1+\frac{2}{3}+\frac{1}{5}\right]- \pileft[-1+\frac{2}{5}-\frac{1}{5}\right] \\[2ex] &= \frac{16\pi}{15} \end{align} </math>"

New page

<math>
x=y^2,
x=1; \text{about x=1}
</math>

<math>
\begin{align}
\pi\int_-1^1\left(1-y^2\right)^2 dy &= \pi\int_-1^1\left(1-2y^2+y^4\right)dy \\[2ex]

&= \pi\left[y-\frac{2}{3}y^3+\frac{y^5}{5}\right]\Bigg|_-1^1 \\[2ex]
&= \pi\left[1+\frac{2}{3}+\frac{1}{5}\right]- \pileft[-1+\frac{2}{5}-\frac{1}{5}\right] \\[2ex]
&= \frac{16\pi}{15}

\end{align}
</math>