91Ó°ÊÓ

Prove that the centroid of a circle is the center of the circle.

Let f(x, y, z) and g(x, y, z) be integrable functions on the rectangular solid $\mathfrak{R}=\left\{\left(x,y,z\right)|a_1â\copyright ½xâ\copyright ½a_2,b_1â\copyright ½yâ\copyright ½b_2,c_1â\copyright ½zâ\copyright ½c_2\right\}$. . Use the definition of the triple integral to prove that :

${∭}_{\mathfrak{R}}\left(f(x,y,z)+g(x,y,z)\right)dV={∭}_{\mathfrak{R}}f(x,y,z)dV+{∭}_{\mathfrak{R}}g(x,y,z)dV.$

Let $R$be the radius of a sphere. Use cylindrical coordinates to set up and evaluate a triple integral proving that the volume of the sphere is$\frac{4}{3}{\mathrm{Ï€¸é}}^3$.

Recall that an annulus is the region between two concentric circles. Prove that the centroid of an annulus is the common center of the two circles.

Let $a<b$and $c<d$be real numbers, and let $R$be the

rectangle in the $xy$-plane defined by

$R=\left\{\right(x,y)âˆ\pounds a≤x≤b\text{and}c≤y≤d\}.$

Prove that role="math" localid="1653851117688" ${âˆ\neg }_{R}dA=(b-a)(d-c)$, what is the relation between $R$and product of$(b-a)(d-c)$?

Let $R$be the radius of the base of a cone and $h$be the height of the cone. Use spherical coordinates to set up and evaluate a triple integral proving that the volume of the cone is $\frac{1}{3}\mathrm{Ï€}{R}^{2}h$.

Let $R$be the radius of a sphere. Use cylindrical coordinates to set up and evaluate a triple integral proving that the volume of the sphere is$\frac{4}{3}{\mathrm{Ï€¸é}}^3$.

To convert from cylindrical to rectangular coordinates:

$x=\_\_\_\_,y=\_\_\_\_,z=\_\_\_\_$

Which of the iterated integrals in Exercises below could correctly

be used to evaluate the double integral:

${∫}_2^6{∫}_1^{4}f(x,y)dxdy$

Identify the quantities determined by the integral expressions in Exercises 3-11. If x and y are both measured in centimeters and $\mathrm{ÒÏ}(x,y)$is a densily funclion in grams per square cenlimeter, give the units of the expression.

$\frac{{âˆ\neg }_{\mathrm{Î\copyright }}x\mathrm{ÒÏ}(x,y)dA}{{âˆ\neg }_{\mathrm{Î\copyright }}\mathrm{ÒÏ}(x,y)dA}\text{and}\frac{{âˆ\neg }_{\mathrm{Î\copyright }}y\mathrm{ÒÏ}(x,y)dA}{{âˆ\neg }_{\mathrm{Î\copyright }}\mathrm{ÒÏ}(x,y)dA}$