Question

A cheese with width = 5cm, length = 10cm and height = 5cm is placed at the origin of xyz axes as shown in the figure. Set up a double integral to determine the volume of the cheese.​

Answer 1

Answer:

5cm+5cm gives 10cm

now, 10cm+10cm gives 20cm

and it is your answer

Answer 2

Since the cheese is a rectangular solid with dimensions 5cm x 10cm x 5cm, its volume is given by:

$V = 5\cdot 10\cdot 5 = 250\text{ cm}^3$

To set up a double integral to determine the volume of the cheese, we can divide the cheese into small rectangular prisms of width $\Delta x$, length $\Delta y$, and height $dz$. The volume of each small rectangular prism is given by:

$\Delta V = \Delta x\cdot\Delta y\cdot dz$

We can then integrate over all the small rectangular prisms to obtain the total volume of the cheese. The limits of integration are 0 to 5 for $x$, 0 to 10 for $y$, and 0 to 5 for $z$. Therefore, the double integral is:

$\iint\limits_{R} dz\,\mathrm{d}A$

where $R$ is the region in the $xy$ plane that corresponds to the base of the cheese. Since the cheese is centered at the origin, this region is simply the rectangle with vertices at [(5/2,-10/2), (-5/2,-10/2), (-5/2,10/2), (5/2,10/2)], or equivalently, at [(2.5,-5), (-2.5,-5), (-2.5,5), (2.5,5)]. Therefore, we can write:

$\begin{aligned} \iint\limits_{R} dz\,\mathrm{d}A &= \int_{-5/2}^{5/2} \int_{-10/2}^{10/2} dz\,dy\,dx \\ &= \int_{-5/2}^{5/2} \int_{-10/2}^{10/2} dz\cdot dy\cdot dx \\ &= \int_{-5/2}^{5/2} \left[z\right]_{-10/2}^{10/2} dy\cdot dx \\ &= \int_{-5/2}^{5/2} (5) dy\cdot dx \\ &= 5\cdot\left[y\right]_{-10/2}^{10/2}\cdot\left[x\right]_{-5/2}^{5/2} \\ &= 5\cdot 10\cdot 5 \\ &= 250\text{ cm}^3 \end{aligned}$

Therefore, the volume of the cheese is 250 cubic centimeters.