Find the volume using disk/washer method of region bound by curves y=e^-x/2, y=ln(9)? The solid is generated when R is revolves around x-axis. The boundaries are ln(9) and 0. I just need help setting up the integral. Thank you.

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Find the volume using disk/washer method of region bound by curves y=e^-x/2, y=ln(9)? The solid is generated when R is revolves around x-axis. The boundaries are ln(9) and 0. I just need help setting up the integral. Thank you.

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Answer 1 · 2018-03-24T02:59:54

See below.

Explanation:

If you look at the graph of $y = ln 9$ $y=ln9$ and $y = e^{- \frac{x}{2}}$ $y=e^[-x/2$ you will notice that the area bounded by $y = ln 9$ $y=ln9$ , from $0$ $0$ to $ln 9$ $ln9$ is a square, and has an area greater than $y = e^{- \frac{x}{2}}$ $y=e^[-x/2$ from $0$ $0$ to $ln 9$ $ln9$ .

So we would need to find $π \int_{0}^{ln 9}$ $piint_[0]^ln9$ ${[ln 9]}^{2}$ $[ln9]^2$ dx - $π \int_{0}^{ln 9}$ $piint_[0]^ln9$ ${[e^{- \frac{x}{2}}]}^{2}]$ $[e^[-x/2]]^2]$ dx

Volume of solid of revolution= $π \int_{a}^{b} [y^{2}] d x$ $piint_a^b[y^2]dx$ , where $y = f [x]$ $y=f[x]$ . Sorry I am unable to show the graphs[ don't know how!]

Leave it to you to evaluate the integrals. Hope this was helpful.

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Find the volume using disk/washer method of region bound by curves y=e^-x/2, y=ln(9)? The solid is generated when R is revolves around x-axis. The boundaries are ln(9) and 0. I just need help setting up the integral. Thank you.

1 Answer

Explanation:

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