Triple Integral: $\dfrac{\sqrt{1-x^2}}{2(1+y)}$

In summary: Addendum: Actually, no. You've got a typo. That first (inside) integral has an upper limit of \sqrt{1 - z^2}, not \sqrt{1 - x^2}. But you've got the idea down.$\displaystyle\int_0^1 \dfrac{z \sqrt{1 - z^2}}{y + 1} dz = \dfrac{1}{y + 1} \int_0^1 z \sqrt{1 - z^2} ~ dz$Let $z = \sin u$, then $dz = \cos u ~ du$.So, $\displaystyle\int_0^1 \dfrac
  • #1
karush
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ok this is a snip from stewards v8 15.6 ex
hopefully to do all 3 here

$\displaystyle\int_0^1\int_{0}^{1}\int_{0}^{\sqrt{1-x^2}}\dfrac{z}{y+1} \,dxdzdy$
so going from the center out but there is no x in the integrand

$\displaystyle\int_0^{\sqrt{1 - x^2}} \dfrac{z}{y + 1}dx =\dfrac{ \sqrt{1 - x^2} z}{y + 1}$
and

$\displaystyle\int_0^1 \dfrac{ \sqrt{1 - x^2} z}{y + 1} \, dz=\frac{\sqrt{1-x^2}}{2\left(1+y\right)} $

kinda maybe so far?
 

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  • #2
karush said:
ok this is a snip from stewards v8 15.6 ex
hopefully to all 3 here

$\displaystyle\int_0^1\int_{0}^{1}\int_{0}^{\sqrt{1-x^2}}\dfrac{z}{y+1} \,dxdzdy$
so going from the center out but there is no x in the integrand

$\displaystyle\int_0^{\sqrt{1 - x^2}} \dfrac{z}{y + 1}dx =\dfrac{ \sqrt{1 - x^2} z}{y + 1}$
and

$\displaystyle\int_0^1 \dfrac{ \sqrt{1 - x^2} z}{y + 1} \, dz=\frac{\sqrt{1-x^2}}{2\left(1+y\right)} $

kinda maybe so far?
You've got it! Keep going...

-Dan

Addendum: Actually, no. You've got a typo. That first (inside) integral has an upper limit of \(\displaystyle \sqrt{1 - z^2}\), not \(\displaystyle \sqrt{1 - x^2}\). But you've got the idea down.
 
  • #3
$\displaystyle\int_0^1\int_{0}^{1}\int_{0}^{\sqrt{1-z^2}}\dfrac{z}{y+1} \,dxdzdy$
so going from the center out but there is no x in the integrand

$\displaystyle\int_0^{\sqrt{1 -z^2}} \dfrac{z}{y + 1}dx =\dfrac{ \sqrt{1 - z^2} z}{y + 1}$
and

$\displaystyle\int_0^1 \dfrac{ \sqrt{1 - z^2} z}{y + 1} \, dz=\frac{\sqrt{1-x^2}}{2\left(1+y\right)} $wait I think I ? on this z thing...
 
  • #4
karush said:
$\displaystyle\int_0^1\int_{0}^{1}\int_{0}^{\sqrt{1-z^2}}\dfrac{z}{y+1} \,dxdzdy$
so going from the center out but there is no x in the integrand

$\displaystyle\int_0^{\sqrt{1 -z^2}} \dfrac{z}{y + 1}dx =\dfrac{ \sqrt{1 - z^2} z}{y + 1}$
and

$\displaystyle\int_0^1 \dfrac{ \sqrt{1 - z^2} z}{y + 1} \, dz=\frac{\sqrt{1-x^2}}{2\left(1+y\right)} $wait I think I ? on this z thing...
Take it step by step. You've got this.
\(\displaystyle \int_0^1 \dfrac{z \sqrt{1 - z^2}}{y + 1} dz = \dfrac{1}{y + 1} \int_0^1 z \sqrt{1 - z^2} ~ dz\)

Now do a trig substitution...

-Dan
 

Related to Triple Integral: $\dfrac{\sqrt{1-x^2}}{2(1+y)}$

What is a triple integral?

A triple integral is a type of integral in multivariable calculus that involves integrating a function over a three-dimensional region. It is represented by three nested integral symbols and is used to calculate volume, mass, and other quantities in three-dimensional space.

What is the formula for a triple integral?

The formula for a triple integral is ∭f(x,y,z) dV, where f(x,y,z) is the function being integrated and dV represents the infinitesimal volume element. It can also be written in terms of the limits of integration as ∭∭∭f(x,y,z) dx dy dz.

What is the purpose of the function in the triple integral?

The function in the triple integral represents the value of a quantity at a given point in three-dimensional space. It is what is being integrated over the region to calculate the total value of the quantity in the region.

Why is there a square root in the function of the triple integral?

The square root in the function of the triple integral is likely there because the region being integrated over has a curved boundary, which requires the use of a different coordinate system. In this case, the function may be representing the distance from the origin to a point on the boundary, which would require the use of a square root.

How do you solve a triple integral?

To solve a triple integral, you first need to set up the limits of integration by determining the boundaries of the three-dimensional region. Then, you can evaluate the integral by using techniques such as substitution or integration by parts. It is important to carefully consider the order of integration and make sure to correctly set up the integral for the given function and region.

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