Let d be the region bounded by the paraboloids; z = 6 - x² - y² and z = x² + y².
write six different triple iterated integrals for the volume of d. evaluate one of the integrals.

∫∫∫1 dV=4\sqrt{3}π

Step-by-step explanation:

From Exercise we have  

z=6-x^{2}-y^{2}

z=x^{2}+y^{2}

we get

2z=6

z=3

x^{2}+y^{2}=3

We use the polar coordinates, we get

x=r cosθ

y=r sinθ

x^{2}+y^{2}&=r^{2}

r^{2}=3

We get at the limits of the variables that well need for our integral

x^{2}+y^{2}≤z≤3

0≤r ≤\sqrt{3}

0≤θ≤2π

Therefore, we get a triple integral

\int \int \int 1\, dV&=\int \int \left(\int_{x^2+y^2}^{3} 1\, dz\right) dA

=\int \int \left(z|_{x^2+y^2}^{3} \right) dA

=\int \int\ \left(3-(x^2+y^2) \right) dA

=\int \int\ \left(3-r^2 \right) dA

=\int_{0}^{2\pi}\int_{0}^{\sqrt{3}} (3-r^2) dr dθ

=3\int_{0}^{2\pi}\int_{0}^{\sqrt{3}}  1 dr dθ-\int_{0}^{2\pi}\int_{0}^{\sqrt{3}} r^2 dr dθ

=3\int_{0}^{2\pi} r|_{0}^{\sqrt{3}}  dθ-\int_{0}^{2\pi} \frac{r^3}{3}|_{0}^{\sqrt{3}}dθ

=3\sqrt{3}\int_{0}^{2\pi} 1 dθ-\sqrt{3}\int_{0}^{2\pi} 1 dθ

=3\sqrt{3} ·2π-\sqrt{3}·2π

=4\sqrt{3}π

We get

∫∫∫1 dV=4\sqrt{3}π

proofs involving isosceles triangles often require special consideration because an isosceles triangle has several distinct properties that do not apply to normal triangles.(more about triangle types) therefore, when you are trying to prove that two triangles are congruent, and one or both triangles, are isosceles you have a few theorems that you can use to make your life easier.

step-by-step explanation:

an isosceles triangle has two congruent sides and two congruent angles. the congruent angles are called the base angles and the other angle is known as the vertex angle. ∠bac and ∠bca are the base angles of the triangle picture on the left. the vertex angle is ∠abc

y/x of point b= y/x of point a

step-by-step explanation:

(i got the same question, this was the correct answer.)