Find volume enclosed by z = 3x^2 + y^2, x + y = 1, x = 0, y = 0, z = 0

[smith1993123]

\(\displaystyle \mbox{a) Find the volume of the solid enclosed by: }\, \)

. . . . .\(\displaystyle z\, =\, 3x^2\, +\, y^2,\, x\, +\, y\, =\, 1,\, x\, =\, 0,\, y\, =\, 0,\,\mbox{ and }\, z\, =\, 0.\)

\(\displaystyle \mbox{b) Evaluate }\, \)

. . . . .\(\displaystyle \displaystyle \int_0^4\, \int_{\sqrt{\strut x\,}}^{2}\, \sqrt{\strut x\, +\, 3y^2\, }\, dy\, dx\, \)

\(\displaystyle \mbox{ by reversing the order of integration.}\)

 

[Deleted member 4993]

\(\displaystyle \mbox{a) Find the volume of the solid enclosed by: }\, \)

. . . . .\(\displaystyle z\, =\, 3x^2\, +\, y^2,\, x\, +\, y\, =\, 1,\, x\, =\, 0,\, y\, =\, 0,\,\mbox{ and }\, z\, =\, 0.\)

\(\displaystyle \mbox{b) Evaluate }\, \)

. . . . .\(\displaystyle \displaystyle \int_0^4\, \int_{\sqrt{\strut x\,}}^{2}\, \sqrt{\strut x\, +\, 3y^2\, }\, dy\, dx\, \)

\(\displaystyle \mbox{ by reversing the order of integration.}\)

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[Prove It]

\(\displaystyle \mbox{a) Find the volume of the solid enclosed by: }\, \)

. . . . .\(\displaystyle z\, =\, 3x^2\, +\, y^2,\, x\, +\, y\, =\, 1,\, x\, =\, 0,\, y\, =\, 0,\,\mbox{ and }\, z\, =\, 0.\)

\(\displaystyle \mbox{b) Evaluate }\, \)

. . . . .\(\displaystyle \displaystyle \int_0^4\, \int_{\sqrt{\strut x\,}}^{2}\, \sqrt{\strut x\, +\, 3y^2\, }\, dy\, dx\, \)

\(\displaystyle \mbox{ by reversing the order of integration.}\)

To help you get started, it helps to visualise what the region looks like. As we know the region is bounded by x = 0, y = 0 and z = 0, we are in the first octant. The rest of the region has an elliptic paraboloid as the upper boundary and a plane as the right hand boundary (it can be thought of as the line y = -x + 1 infinitely stacked on top of itself).

So the upper and lower boundaries are \(\displaystyle \displaystyle \begin{align*} 0 \leq z \leq 3\,x^2 + y^2 \end{align*}\).

In the north-south direction, we have boundaries \(\displaystyle \displaystyle \begin{align*} 0 \leq y \leq -x + 1 \end{align*}\).

The largest value of x in the x-y plane is given where y = 0 (so x = 1). Thus \(\displaystyle \displaystyle \begin{align*} 0 \leq x \leq 1 \end{align*}\).

Thus the volume is given by \(\displaystyle \displaystyle \begin{align*} V = \int_0^1{\int_0^{-x + 1}{\int_0^{3\,x^2 + y^2}{\,\mathrm{d}x}\,\mathrm{d}y}\,\mathrm{d}z} \end{align*}\)

Now do the integration...


For part (b), you should start by drawing a sketch of your region.