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Can someone walk me through with evaluating, I've done it over and over and yet I am not getting the same answer as the solutions.
[tex] \int\limits^r_{\sqrt{r^2- \frac{h^2}{4}} ^[/tex] [tex]4 \pi x \sqrt{r^2-x^2} dx[/tex]

Respuesta :

[tex]\displaystyle\int_{\sqrt{r^2-\frac{h^2}4}}^r4\pi x\sqrt{r^2-x^2}\,\mathrm dx[/tex]

Let [tex]u=r^2-x^2[/tex], so that [tex]-\dfrac{\mathrm du}2=x\,\mathrm dx[/tex]. Then when [tex]x=\sqrt{r^2-\dfrac{h^2}4}[/tex], you have [tex]u=\dfrac{h^2}4[/tex]; and when [tex]x=r[/tex], you have [tex]u=0[/tex].

So the integral is equivalent to

[tex]\displaystyle-\dfrac12\int_{\frac{h^2}4}^04\pi\sqrt u\,\mathrm du[/tex]
[tex]\displaystyle2\pi\int_0^{\frac{h^2}4}\sqrt{u}\,\mathrm du[/tex]
[tex]2\pi\times\dfrac23u^{3/2}\bigg|_{u=0}^{u=h^2/4}[/tex]
[tex]\dfrac{4\pi}3\left(\dfrac{h^2}4\right)^{3/2}[/tex]
[tex]\dfrac\pi6h^3[/tex]

assuming [tex]h>0[/tex], which is probably the case since the integral appears to represent one that computes the volume of a solid of revolution.
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