Re: Calculus
pamsandhu said:
Evaluate the integral \iiint\limits_{ydV}, where V is the solid lying below the plane x+y+z =8 and above the region in the x-y plane bounded by the curves y=1, x=0 and x=\sqrt{y}.
Lying above the x-y plane and below the plane $\displaystyle \begin{align*} x + y + z = 8 \end{align*}$ means the z limits are $\displaystyle \begin{align*} 0 \leq z \leq 8 - x- y \end{align*}$.
If you are bounded by $\displaystyle \begin{align*} y = 1, \, x = 0 \end{align*}$ and $\displaystyle \begin{align*} x = \sqrt{y} \end{align*}$ then $\displaystyle \begin{align*} 0 \leq x \leq \sqrt{y} \end{align*}$ and $\displaystyle \begin{align*} 0 \leq y \leq 1 \end{align*}$ (draw a sketch of this region to verify). Thus
$\displaystyle \begin{align*} \int{\int{\int_D{y\,\mathrm{d}V}}} &= \int_0^{1}{\int_0^{\sqrt{y}}{\int_0^{8-x-y}{y\,\mathrm{d}z}\,\mathrm{d}x}\,\mathrm{d}y} \\ &= \int_0^1{ \int_0^{\sqrt{y}}{\left[ y\,z \right]_0^{8-x-y} \,\mathrm{d}x} \,\mathrm{d}y } \\ &= \int_0^1{\int_0^{\sqrt{y}}{ y\left( 8 - x- y \right) \,\mathrm{d}x}\,\mathrm{d}y} \\ &= \int_0^1{\left[ y\left( 8\,x - \frac{x^2}{2} - x\,y \right) \right]_0^{\sqrt{y}}\,\mathrm{d}y} \\ &= \int_0^1{ y\left( 8\,\sqrt{y} - \frac{y}{2} - y\,\sqrt{y} \right) \,\mathrm{d}y } \\ &= \int_0^1{ \left( 8\,y^{\frac{3}{2}} - \frac{y^2}{2} - y^{\frac{5}{2}} \right) \,\mathrm{d}y } \\ &= \left[ \frac{16}{5}\,y^{\frac{5}{2}} - \frac{y^3}{6} - \frac{2}{7}\,y^{\frac{7}{2}} \right]_0^1 \\ &= \frac{16}{5} - \frac{1}{6} - \frac{2}{7} \\ &= \frac{577}{210} \end{align*}$
