Question #219833

Use a triple integral to determine the volume of the region bounded by

z

=

x

2

+

y

2

and

z

=

x

2

+

y

2

in 1st octant.



1
Expert's answer
2021-07-23T09:54:24-0400

By equating the coordinates z,z, we get the following equation:


x2+y2=x2+y2\sqrt{x^2+y^2}=x^2+y^2

x2+y2=0 or x2+y2=1x^2+y^2=0\ or\ x^2+y^2=1

Using the cylindrical coordinates

V=EdV=V=\int\int\int_EdV=

=0π/201r2rrdzdrdθ=\displaystyle\int_{0}^{\pi/2}\displaystyle\int_{0}^{1}\displaystyle\int_{r^2}^{r}rdzdrd\theta

=0π/201(r2r3)drdθ=\displaystyle\int_{0}^{\pi/2}\displaystyle\int_{0}^{1}(r^2-r^3)drd\theta

=0π/2[r33r44]10dθ=\displaystyle\int_{0}^{\pi/2}\bigg[\dfrac{r^3}{3}-\dfrac{r^4}{4}\bigg]\begin{matrix} 1 \\ 0 \end{matrix}d\theta

=112[θ]π/20=π24(units3)=\dfrac{1}{12}\big[\theta\big]\begin{matrix} \pi/2 \\ 0 \end{matrix}=\dfrac{\pi}{24}(units^3)

V=π24V=\dfrac{\pi}{24} cubic units



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