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The question itself is:

A diving pool that is 4 meters deep, and full of water, has on one of its walls a circular viewing window, tangent to the bottom of the pool, with a radius of .5 meters. Find the force on this window.

I have the integral $\int_0^1 1000\cdot9.8\cdot(4-y)\sqrt{0.25-y^2}\,\mathrm dx$, and I have absolutely no idea how to work with it. I've tried multiple different approaches and every time I simply cannot do it. Any help would be appreciated.

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    $\begingroup$ Note that for $y>1/2$, $0.25-y^2<0$. Maybe the upper bound of the integral is not $1$? $\endgroup$ – Stop hurting Monica Sep 4 at 19:01
  • $\begingroup$ Which approaches have you tried? Can you find $\int 4 \sqrt{0.25 - y^2}\,dy$ and $\int y \sqrt{0.25 - y^2}\,dy$? $\endgroup$ – Matthew Leingang Sep 4 at 19:08
  • $\begingroup$ @MatthewLeingang I have tried this, but get stuck there as well. I'm severely struggling with this problem on all fronts. $\endgroup$ – Revellion Sep 4 at 19:10
  • $\begingroup$ “If you can't solve a problem, then there is an easier problem you also can't solve: find it” (Polyà). It sounds like you want to try to find those two antiderivatives. The second is easier than the first. $\endgroup$ – Matthew Leingang Sep 4 at 19:13
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    $\begingroup$ @Jean-ClaudeArbaut: My certainly led me to reread all of How to Solve It twice and I couldn't find it either way. Then I found this question. Apparently Conway attributed it to Polyà with can't. The way I always thought about it, the idea is simplify the problem as much as you can, without removing the part that makes it difficult. So you can't find $\int\sqrt{0.25-y^2}\,dy$? What about $\int\sqrt{1-y^2}\,dy$ isntead? That's easier, but maintains the essential hardness of the first. $\endgroup$ – Matthew Leingang Sep 4 at 22:51
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Hint

Write it as

$(1000•9.8•4)-(1000•9.8•y )(0.25-y^2)^{0.5}$

then use the substitution $0.25-y^2=y$ for the second part and integrate the first part manually.

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  • $\begingroup$ How did you separate the integral into those two parts? the y is being subtracted from 4, and there it looks like you completely took it apart. $\endgroup$ – Revellion Sep 4 at 19:24
  • $\begingroup$ I multiplied the constants together and then multiplied them by $y$. If you have $a(b-c)$ you can write it as $ab-ac$ $\endgroup$ – Sina Babaei Zadeh Sep 4 at 20:46
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Hint: Split the integrand into two parts: the first a constant multiple of $\sqrt{0.25-y^2},$ and the second a constant multiple of $-y\sqrt{0.25-y^2}.$ You may do the first by using the substitution $$y=\frac12\sin\phi.$$ To do the second, use the substitution $u=0.25-y^2.$

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