Volume for Triple Integral

[harpazo]

Use a triple integral to find the volume of the solid bounded by the graphs of the equations.

x = 4 - y^2, z = 0, z = x

I need help setting up the triple integral for the volume. I will do the rest.

 

[Prove It]

Use a triple integral to find the volume of the solid bounded by the graphs of the equations.

x = 4 - y^2, z = 0, z = x

I need help setting up the triple integral for the volume. I will do the rest.

Well we can clearly see that $\displaystyle \begin{align*} 0 \leq z \leq x \end{align*}$, so that also means that $\displaystyle \begin{align*} x \geq 0 \end{align*}$. Now since x is bounded above by $\displaystyle \begin{align*} x = 4 - y^2 \end{align*}$, a parabola with y intercepts -2 and 2, that means that $\displaystyle \begin{align*} 0 \leq x \leq 4 - y^2 \end{align*}$ and $\displaystyle \begin{align*} -2 \leq y \leq 2 \end{align*}$. So the volume is

$\displaystyle \begin{align*} V &= \int_{-2}^2{\int_0^{4 - y^2}{\int_0^x{\,\mathrm{d}z}\,\mathrm{d}x}\,\mathrm{d}y} \end{align*}$

Can you go from here?

 

[MarkFL]

As a review, there are several other ways to find the given volume. From Calc I, recall "Volumes By Slicing."

If we use slices perpendicular to the $xy$-plane, and parallel to the $x$-axis, we find an arbitrary slice is a right-isosceles triangle, whose legs are $x$ in length. And so the volume of such a slice is:

\(\displaystyle dV=\frac{1}{2}x^2\,dy=\frac{1}{2}(4-y^2)^2\,dy=\frac{1}{2}(16-8y^2+y^4)\,dy\)

Summing the slices via integration (and using the even-function rule), we find:

\(\displaystyle V=\int_0^2 16-8y^2+y^4\,dy=\left[16y-\frac{8}{3}y^3+\frac{1}{5}y^5\right]_0^2=\frac{1}{15}\left(240(2)-40(8)+3(32)\right)=\frac{256}{15}\)

If we use slices perpendicular to the $xy$-plane, and parallel to the $y$-axis, we find an arbitrary slice is a rectangle, whose base is $2y$ and whose height is $x$. And so the volume of such a slice is:

\(\displaystyle dV=2xy\,dx=2x\sqrt{4-x}\,dx\)

Summing the slices via integration, we find:

\(\displaystyle V=2\int_0^4 x\sqrt{4-x}\,dx\)

Use the substitution:

\(\displaystyle u=4-x\implies du=-dx\)

\(\displaystyle V=2\int_0^4 (4-u)u^{\frac{1}{2}}\,du=2\int_0^4 4u^{\frac{1}{2}}-u^{\frac{3}{2}}\,du=2\left[\frac{8}{3}u^{\frac{3}{2}}-\frac{2}{5}u^{\frac{5}{2}}\right]_0^4=\frac{2}{15}\left(40(8)-6(32)\right)=\frac{256}{15}\)

If we use slices that are parallel to the $xy$-plane, then we find these slices are parabolic segments. So, we need to find the area of such a segment:

\(\displaystyle A=2\int_0^{\sqrt{h}} h-x^2\,dx=2\left[hx-\frac{1}{3}x^3\right]_0^{\sqrt{h}}=\frac{4}{3}h^{\frac{3}{2}}\)

Thus, the volume of an arbitrary slice is:

\(\displaystyle dV=\frac{4}{3}(4-z)^{\frac{3}{2}}\,dz\)

Summing the slices via integration, we find:

\(\displaystyle V=\frac{4}{3}\int_0^4 (4-z)^{\frac{3}{2}}\,dz\)

Let:

\(\displaystyle u=4-z\implies du=-dz\)

\(\displaystyle V=\frac{4}{3}\int_0^4 u^{\frac{3}{2}}\,du=\frac{8}{15}\left[u^{\frac{5}{2}}\right]_0^4=\frac{256}{15}\)

Double integral - vertical slices:

\(\displaystyle V=2\int_0^4 x\int_0^{\sqrt{4-x}} \,dy\,dx=2\int_0^4 x\sqrt{4-x}\,dx=\frac{256}{15}\)

Double integral - horizontal slices:

\(\displaystyle V=2\int_0^2\int_0^{4-y^2} x\,dx\,dy=\int_0^2 (4-y^2)^2\,dy=\frac{256}{15}\)

 

[harpazo]

Well we can clearly see that $\displaystyle \begin{align*} 0 \leq z \leq x \end{align*}$, so that also means that $\displaystyle \begin{align*} x \geq 0 \end{align*}$. Now since x is bounded above by $\displaystyle \begin{align*} x = 4 - y^2 \end{align*}$, a parabola with y intercepts -2 and 2, that means that $\displaystyle \begin{align*} 0 \leq x \leq 4 - y^2 \end{align*}$ and $\displaystyle \begin{align*} -2 \leq y \leq 2 \end{align*}$. So the volume is

$\displaystyle \begin{align*} V &= \int_{-2}^2{\int_0^{4 - y^2}{\int_0^x{\,\mathrm{d}z}\,\mathrm{d}x}\,\mathrm{d}y} \end{align*}$

Can you go from here?

Yes, I can go from here. You did not explain why x is greater than or equal to 0. I want to understand this a little more.

 

[harpazo]

As a review, there are several other ways to find the given volume. From Calc I, recall "Volumes By Slicing."

If we use slices perpendicular to the $xy$-plane, and parallel to the $x$-axis, we find an arbitrary slice is a right-isosceles triangle, whose legs are $x$ in length. And so the volume of such a slice is:

\(\displaystyle dV=\frac{1}{2}x^2\,dy=\frac{1}{2}(4-y^2)^2\,dy=\frac{1}{2}(16-8y^2+y^4)\,dy\)

Summing the slices via integration (and using the even-function rule), we find:

\(\displaystyle V=\int_0^2 16-8y^2+y^4\,dy=\left[16y-\frac{8}{3}y^3+\frac{1}{5}y^5\right]_0^2=\frac{1}{15}\left(240(2)-40(8)+3(32)\right)=\frac{256}{15}\)

If we use slices perpendicular to the $xy$-plane, and parallel to the $y$-axis, we find an arbitrary slice is a rectangle, whose base is $2y$ and whose height is $x$. And so the volume of such a slice is:

\(\displaystyle dV=2xy\,dx=2x\sqrt{4-x}\,dx\)

Summing the slices via integration, we find:

\(\displaystyle V=2\int_0^4 x\sqrt{4-x}\,dx\)

Use the substitution:

\(\displaystyle u=4-x\implies du=-dx\)

\(\displaystyle V=2\int_0^4 (4-u)u^{\frac{1}{2}}\,du=2\int_0^4 4u^{\frac{1}{2}}-u^{\frac{3}{2}}\,du=2\left[\frac{8}{3}u^{\frac{3}{2}}-\frac{2}{5}u^{\frac{5}{2}}\right]_0^4=\frac{2}{15}\left(40(8)-6(32)\right)=\frac{256}{15}\)

If we use slices that are parallel to the $xy$-plane, then we find these slices are parabolic segments. So, we need to find the area of such a segment:

\(\displaystyle A=2\int_0^{\sqrt{h}} h-x^2\,dx=2\left[hx-\frac{1}{3}x^3\right]_0^{\sqrt{h}}=\frac{4}{3}h^{\frac{3}{2}}\)

Thus, the volume of an arbitrary slice is:

\(\displaystyle dV=\frac{4}{3}(4-z)^{\frac{3}{2}}\,dz\)

Summing the slices via integration, we find:

\(\displaystyle V=\frac{4}{3}\int_0^4 (4-z)^{\frac{3}{2}}\,dz\)

Let:

\(\displaystyle u=4-z\implies du=-dz\)

\(\displaystyle V=\frac{4}{3}\int_0^4 u^{\frac{3}{2}}\,du=\frac{8}{15}\left[u^{\frac{5}{2}}\right]_0^4=\frac{256}{15}\)

Double integral - vertical slices:

\(\displaystyle V=2\int_0^4 x\int_0^{\sqrt{4-x}} \,dy\,dx=2\int_0^4 x\sqrt{4-x}\,dx=\frac{256}{15}\)

Double integral - horizontal slices:

\(\displaystyle V=2\int_0^2\int_0^{4-y^2} x\,dx\,dy=\int_0^2 (4-y^2)^2\,dy=\frac{256}{15}\)

You are truly gifted. Thank you very much.

 

[Prove It]

Yes, I can go from here. You did not explain why x is greater than or equal to 0. I want to understand this a little more.

Yes I did, $\displaystyle \begin{align*} 0 \leq z \leq x \end{align*}$ literally says $\displaystyle \begin{align*} 0 \leq x \end{align*}$!

 

[harpazo]

Yes I did, $\displaystyle \begin{align*} 0 \leq z \leq x \end{align*}$ literally says $\displaystyle \begin{align*} 0 \leq x \end{align*}$!

Sorry but I do not get it.

 

[skeeter]

Maybe a visual will help ...

Dark green surface is the plane $z=x$.

Light green surface is the plane $z=0$.

Purple surface is the parabola $x=4-y^2$

purple "wedge" in octants I and IV between the two planes is the solid of interest.

 

[harpazo]

Thank you everyone.