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I found the derivation of finding the Volume of a cylinder using triple integral with Cartesian coordinates. However, I cant seem to find out how they got '2h' in here. Am I missing something?

The derivation can be found in the image below:

$$\iiint_{E}\,dV=\int_{-a}^{a} \int_{-\sqrt{a^2-x^2}}^{\sqrt{a^2-x^2}} \int_0^h\,1 dzdydx=\int_{-a}^{a} 2h \sqrt{a^2-x^2}\,dx=2h\dfrac12\pi a^2=\pi a^2h $$

Thank you so much.

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  • $\begingroup$ $h$ is from the last integral, and $2$ from the middle one: $\sqrt{a^2-x^2}-(-\sqrt{a^2-x^2})=2\sqrt{a^2-x^2}$ $\endgroup$
    – Sine of the Time
    Jan 27 at 19:38

1 Answer 1

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The steps are the following:$$\int\limits_0^h1dz=h$$and $$\int\limits_{-\sqrt{a^2-x^2}}^{\sqrt{a^2-x^2}}hdy=h\left(y\rvert_{-\sqrt{a^2-x^2}}^{\sqrt{a^2-x^2}}\right)=h\left({\sqrt{a^2-x^2}}-({-\sqrt{a^2-x^2}})\right)=2h\sqrt{a^2-x^2}$$

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