How to Set Up a Triple Integral for Volume in the First Octant?

[harpazo]

Use a triple integral to find the volume of the solid bounded by the graphs of the equations.

z = 4 - x^2, y = 4 - x^2, first octant

I need help setting up the triple integral for the volume. I will do the rest.

 

[Prove It]

Use a triple integral to find the volume of the solid bounded by the graphs of the equations.

z = 4 - y^2, y = 4 - x^2, first octant

I need help setting up the triple integral for the volume. I will do the rest.

Well as you know it's bounded in the first octant, that means all variables are nonnegative, so the lower bound for each variable is 0.

So we could have $\displaystyle \begin{align*} 0 \leq z \leq 4 - y^2 \end{align*}$.

Now as we know the upper bound for y is $\displaystyle \begin{align*} y = 4 - x^2 \end{align*}$, a parabola with x intercepts -2 and 2, that means $\displaystyle \begin{align*} 0 \leq y \leq 4 - x^2 \end{align*}$ and $\displaystyle \begin{align*} 0 \leq x \leq 2 \end{align*}$. Thus the volume integral is

$\displaystyle \begin{align*} V &= \int_0^2{\int_0^{4 - x^2}{\int_0^{4 - y^2}{\,\mathrm{d}z}\,\mathrm{d}y}\,\mathrm{d}x} \\ &= \int_0^2{\int_0^{4 - x^2}{\left[ z \right] _0^{4 - y^2}\,\mathrm{d}y}\,\mathrm{d}x} \\ &= \int_0^2{\int_0^{4 - x^2}{\left( 4 - y^2 - 0 \right) \,\mathrm{d}y}\,\mathrm{d}x} \\ &= \int_0^2{ \int_0^{4 - x^2}{ \left( 4 - y^2 \right) \,\mathrm{d}y } \,\mathrm{d}x} \\ &= \int_0^2{ \left[ 4\,y - \frac{1}{3}\,y^3 \right]_0^{4 - x^2} \,\mathrm{d}x } \\ &= \int_0^2{ \left\{ \left[ 4 \left( 4 - x^2 \right) - \frac{1}{3} \left( 4 - x^2 \right) ^3 \right] - \left[ 4 \left( 0 \right) - \frac{1}{3} \left( 0 \right) ^3 \right] \right\} \,\mathrm{d}x } \\ &= \int_0^2{ \left[ 16 - 4\,x^2 - \frac{1}{3} \left( 64 - 48\,x^2 + 12\,x^4 - x^6 \right) \right] \,\mathrm{d}x } \\ &= \int_0^2{ \left( 16 - 4\,x^2 - \frac{64}{3} + 16\,x^2 - 4\,x^4 + \frac{1}{3}\,x^6 \right) \,\mathrm{d}x } \\ &= \int_0^2{ \left( \frac{1}{3}\,x^6 - 4\,x^4 + 12\,x^2 - \frac{18}{3} \right) \,\mathrm{d}x } \\ &= \left[ \frac{1}{21}\,x^7 - \frac{4}{5}\,x^5 + 4\,x^3 - \frac{18}{3}\,x \right] _0^2 \\ &= \left[ \frac{1}{21}\left( 2 \right) ^7 - \frac{4}{5} \left( 2 \right) ^5 + 4 \left( 2 \right) ^3 - \frac{18}{3} \left( 2 \right) \right] - \left[ \frac{1}{21} \left( 0 \right) ^7 - \frac{4}{5} \left( 0 \right) ^5 + 4 \left( 0 \right) ^3 - \frac{18}{3} \left( 0 \right) \right] \\ &= \frac{128}{21} - \frac{128}{5} + 32 - \frac{36}{3} - 0 \\ &= \frac{640}{105} - \frac{2688}{105} + \frac{3360}{105} - \frac{1260}{105} \\ &= \frac{52}{105} \end{align*}$

 

[harpazo]

Well as you know it's bounded in the first octant, that means all variables are nonnegative, so the lower bound for each variable is 0.

So we could have $\displaystyle \begin{align*} 0 \leq z \leq 4 - y^2 \end{align*}$.

Now as we know the upper bound for y is $\displaystyle \begin{align*} y = 4 - x^2 \end{align*}$, a parabola with x intercepts -2 and 2, that means $\displaystyle \begin{align*} 0 \leq y \leq 4 - x^2 \end{align*}$ and $\displaystyle \begin{align*} 0 \leq x \leq 2 \end{align*}$. Thus the volume integral is

$\displaystyle \begin{align*} V &= \int_0^2{\int_0^{4 - x^2}{\int_0^{4 - y^2}{\,\mathrm{d}z}\,\mathrm{d}y}\,\mathrm{d}x} \\ &= \int_0^2{\int_0^{4 - x^2}{\left[ z \right] _0^{4 - y^2}\,\mathrm{d}y}\,\mathrm{d}x} \\ &= \int_0^2{\int_0^{4 - x^2}{\left( 4 - y^2 - 0 \right) \,\mathrm{d}y}\,\mathrm{d}x} \\ &= \int_0^2{ \int_0^{4 - x^2}{ \left( 4 - y^2 \right) \,\mathrm{d}y } \,\mathrm{d}x} \\ &= \int_0^2{ \left[ 4\,y - \frac{1}{3}\,y^3 \right]_0^{4 - x^2} \,\mathrm{d}x } \\ &= \int_0^2{ \left\{ \left[ 4 \left( 4 - x^2 \right) - \frac{1}{3} \left( 4 - x^2 \right) ^3 \right] - \left[ 4 \left( 0 \right) - \frac{1}{3} \left( 0 \right) ^3 \right] \right\} \,\mathrm{d}x } \\ &= \int_0^2{ \left[ 16 - 4\,x^2 - \frac{1}{3} \left( 64 - 48\,x^2 + 12\,x^4 - x^6 \right) \right] \,\mathrm{d}x } \\ &= \int_0^2{ \left( 16 - 4\,x^2 - \frac{64}{3} + 16\,x^2 - 4\,x^4 + \frac{1}{3}\,x^6 \right) \,\mathrm{d}x } \\ &= \int_0^2{ \left( \frac{1}{3}\,x^6 - 4\,x^4 + 12\,x^2 - \frac{18}{3} \right) \,\mathrm{d}x } \\ &= \left[ \frac{1}{21}\,x^7 - \frac{4}{5}\,x^5 + 4\,x^3 - \frac{18}{3}\,x \right] _0^2 \\ &= \left[ \frac{1}{21}\left( 2 \right) ^7 - \frac{4}{5} \left( 2 \right) ^5 + 4 \left( 2 \right) ^3 - \frac{18}{3} \left( 2 \right) \right] - \left[ \frac{1}{21} \left( 0 \right) ^7 - \frac{4}{5} \left( 0 \right) ^5 + 4 \left( 0 \right) ^3 - \frac{18}{3} \left( 0 \right) \right] \\ &= \frac{128}{21} - \frac{128}{5} + 32 - \frac{36}{3} - 0 \\ &= \frac{640}{105} - \frac{2688}{105} + \frac{3360}{105} - \frac{1260}{105} \\ &= \frac{52}{105} \end{align*}$

Fabulous work but the textbook answer is 256/15.

 

[MarkFL]

Fabulous work but the textbook answer is 256/15.

I think you are confusing this problem with another you posted. However, when I ask W|A to evaluate the triple integral set up by Prove It, it spits out:

$$I=\frac{64}{35}$$

 

[harpazo]

I think you are confusing this problem with another you posted. However, when I ask W|A to evaluate the triple integral set up by Prove It, it spits out:

$$I=\frac{64}{35}$$

I just saw my typo. See my edited question.

z = 4 - x^2 not z = 4 - y^2

 

[harpazo]

I also plugged into W.A. and the answer given is 64/35 but the textbook answer is 256/15.