Harpazo said: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.
Prove It said: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 said:Fabulous work but the textbook answer is 256/15.
MarkFL said: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:
\(\displaystyle I=\frac{64}{35}\)
Triple integral volume is a mathematical concept used to calculate the volume of a three-dimensional object. It involves integrating a function over a three-dimensional region, using three sets of variables to represent each dimension.
To set up a triple integral for volume, you need to define the limits of integration for each of the three variables (x, y, and z) and identify the function to be integrated. The limits of integration are typically the boundaries of the three-dimensional region.
Triple integral volume calculates the amount of space enclosed by a three-dimensional object, while surface area calculates the amount of space taken up by the surface of that object. In other words, volume measures the interior space, while surface area measures the exterior space.
Triple integral volume has various real-world applications, such as calculating the volume of a water tank, determining the amount of medicine in a pill, or calculating the amount of air in a room. It is also used in engineering and physics to calculate the volume of complex shapes and objects.
One limitation of using triple integral volume is that it can only be applied to objects with a defined boundary. Additionally, the calculation can become complex and time-consuming for irregularly shaped objects. It also assumes that the object has a constant density, which may not always be the case in real-world situations.