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Using disks or washers, find the volume of the solid obtained by rotating the region bounded by the curves $y=x^2$, $x=5$, and $y=0$ about the $x$-axis.

How do I solve this when given rotating region about the x-axis?

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Your bounds of integration are going to be from $0$ to $5$. Your radius function is going to be $r(x)=x^2$. Now, just plug it into the formula for the disk method which is $V=\pi\int_{a}^{b}\left[r(x)\right]^2\,dx$ and you're good to go:

$$ V=\pi\int_{0}^{5}(x^2)^2\,dx= \pi\int_{0}^{5}x^4\,dx=\pi\frac{x^5}{5}\bigg|_{0}^{5}=\pi\left(\frac{5^5}{5}-\frac{0}{5}\right)=625\pi. $$

Here is a rough picture of what you are doing:

enter image description here

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Using the formula $V = \pi \int_a^b f^2(x) dx$ , the volume you are looking for is the volume gerenated by the curve between $x=0$ and $x=5$, so the volume is

$V = \pi \int_0^5 x^4 dx = 625 \pi$

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  • $\begingroup$ This answer adds nothing new, and came later that the more detailed answer above. $\endgroup$
    – Ramanujan
    Feb 10 '19 at 23:44
  • $\begingroup$ @ViktorGlombik I was the first one posting the solution, check out the answer time $\endgroup$
    – JoseSquare
    Feb 10 '19 at 23:49

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