PullandTwist said:
I'm not getting the right answer... why?
To find the volume of a region D bounded above by $\displaystyle \begin{align*} z = f_1 \left( x, y \right) \end{align*}$ and below by $\displaystyle \begin{align*} z = f_2 \left( x, y \right) \end{align*}$, you need to do
$\displaystyle \begin{align*} V = \int{\int_D{\left[ f_1 \left( x, y \right) - f_2 \left( x, y \right) \right]\,\mathrm{d}V}} \end{align*}$
Here $\displaystyle \begin{align*} f_1 \left( x, y \right) = \sqrt{1 - x^2 - y^2} \end{align*}$ and $\displaystyle \begin{align*} f_2 \left( x, y \right) = 1 - \sqrt{1 - x^2 - y^2} \end{align*}$. You have correctly established that the intersection of the two spheres is the circle at $\displaystyle \begin{align*}z = \frac{1}{2} \end{align*}$ of radius $\displaystyle \begin{align*} \frac{\sqrt{3}}{2} \end{align*}$ centred at $\displaystyle \begin{align*} (x,y) = (0,0) \end{align*}$, and since each cross section is a full circle, that means $\displaystyle \begin{align*} 0 \leq r \leq \frac{\sqrt{3}}{2} \end{align*}$ and $\displaystyle \begin{align*} 0 \leq \theta \leq 2\pi \end{align*}$. So your volume is calculated by
$\displaystyle \begin{align*} \int{\int_D{ \left[ \sqrt{1 - x^2 - y^2} - \left( 1 - \sqrt{1 - x^2 - y^2} \right) \right]\,\mathrm{d}V }} &= \int_0^{2\pi}{\int_0^{\frac{\sqrt{3}}{2}}{ \left[ \sqrt{1 - r^2} - \left( 1 - \sqrt{1 - r^2} \right) \right] \, r\, \mathrm{d}r }\,\mathrm{d}\theta} \\ &= \int_0^{2\pi}{\int_0^{\frac{\sqrt{3}}{2}}{\left( 2\,\sqrt{1 - r^2} - 1 \right) \, r \, \mathrm{d}r }\,\mathrm{d}\theta} \\ &= \int_0^{2\pi}{ -\frac{1}{2} \int_0^{\frac{\sqrt{3}}{2}}{ \left( 2\,\sqrt{1 - r^2} - 1 \right) \left( -2r \right) \,\mathrm{d}r } \,\mathrm{d}\theta } \end{align*}$
Now let $\displaystyle \begin{align*} u = 1 - r^2 \implies \mathrm{d}u = -2r \,\mathrm{d}r \end{align*}$ and note that $\displaystyle \begin{align*} u(0) = 1 \end{align*}$ and $\displaystyle \begin{align*} u \left( \frac{\sqrt{3}}{2} \right) = \frac{1}{4} \end{align*}$, the integral becomes
$\displaystyle \begin{align*} \int_0^{2\pi}{-\frac{1}{2} \int_0^{\frac{\sqrt{3}}{2}}{ \left( 2\,\sqrt{1 - r^2} - 1 \right) \left( -2r \right) \,\mathrm{d}r }\,\mathrm{d}\theta} &= \int_0^{2\pi}{-\frac{1}{2}\int_1^{\frac{1}{4}}{ \left( 2\,\sqrt{u} - 1 \right) \, \mathrm{d}u }\,\mathrm{d}\theta} \\ &= \int_0^{2\pi}{ \frac{1}{2} \int_{\frac{1}{4}}^1{ \left( 2u^{\frac{1}{2}} - 1 \right) \,\mathrm{d}u } \,\mathrm{d}\theta } \\ &= \int_0^{2\pi}{\int_{\frac{1}{4}}^1{ \left( u^{\frac{1}{2}} - \frac{1}{2} \right) \,\mathrm{d}u }\,\mathrm{d}\theta} \\ &= \int_0^{2\pi}{\left[ \frac{2}{3}u^{\frac{3}{2}} - \frac{1}{2}u \right] _{\frac{1}{4}}^1\,\mathrm{d}\theta} \\ &= \int_0^{2\pi}{\left[ \frac{2}{3} \left( 1 \right) ^{\frac{3}{2}} - \frac{1}{2} \left( 1 \right) \right] - \left[ \frac{2}{3} \left( \frac{1}{4} \right) ^{\frac{3}{2}} - \frac{1}{2} \left( \frac{1}{4} \right) \right] \,\mathrm{d}\theta} \\ &= \int_0^{2\pi}{\left( \frac{2}{3} - \frac{1}{2} \right) - \left( \frac{1}{12} - \frac{1}{8} \right) \,\mathrm{d}\theta} \\ &= \int_0^{2\pi}{ \frac{1}{6} - \left( -\frac{1}{24} \right) \,\mathrm{d}\theta } \\ &= \int_0^{2\pi}{ \frac{5}{24} \,\mathrm{d}\theta } \\ &= \frac{5}{24} \left[ \theta \right]_0^{2\pi} \\ &= \frac{5}{24} \left( 2\pi - 0 \right) \\ &= \frac{5\pi}{12} \end{align*}$
It appears your problem was that you ADDED the two functions in the integrand, when you should have SUBTRACTED.
