triple intergral

Question

Evaluate

the triple integral of x^2 where D is the region inside the cylinder x^2+y^2=9 which is bounded below by the plane z=0 and bounded above by the plane 4x+4y+z=19
I know the limits of the z part of the integral are z=0 and z=19-4x-4y

Answer 1

Here is a summary and outline of the process and the procedure per the integral calculator on the website.

A lot of the details and steps are skipped and left for you to check and verify.

 

Integration shall be in the order dz dy dx , inside - out.

 

Specifically,   triple integral  x^2 dz dy dx with the 

 

limits of integration: z = 0 to 19-yx-4y

                             y = -sqrt(9-x^2) to sqrt(9-x^2)

                             x = -3 to 3,

 

 per the base of the cylinder with radius 3 

 

z integrates out to 19-4x-4y

 

The resulting double integral:

 

  x^2(19-4x-4y) dy dx =

 

Integrating with respect to  y:

 

  integral  [(19x^2 - 4x^3)y - 2x^2y^2 ] for y = -sqrt(9-x^2) to sqrt(9-x^2)

 

 When plugging in these limits of integration, the y^2 term cancels because of -sqrt(9-x^2)

 which changes the sign and gets subtracted, the sign changes twice.

 

The final integral is  2(19x^2-4x^3)(9-x^2)^1/2

 

Next, distributes sqrt(9-x^2) = (9-x^2)^(1/2) over (19x^2 - 4x^3) and applies linearity.

The two integrals are:

 

4 * integral x^3(9-x^2)^(1/2) dx     and   19 integral x^2 * (9 - x^2)^1/2 dx

 

[---- call this location BETA, so we can return to this location at a later time ---]

 

The integral on the right begins by rewriting x^2 = x^2 - 9 + 9 and splitting as:

 

     9 integral (9-x^2)^1/2 dx   and integral (9-x^2)^(3/2)

 

[----- call this location ALPHA so we can return here after the next step]

 

Again, this right integral begins with the trig substitution x = 3*sinu --> u = arcsin(x/3) and dx = 2 cosu du

 

This integral becomes  3 integral cosu * (9 - 9sinu^2)^(3/2) =

 

                                 3 integral cosu * (9cos^3) <---- trig substitution cosx^2 = 1 - sinx^2 , then resolving exponent

 

                                3 integral cos u^4

 

Next, the reduction formula

   integral cos t ^N dt = (n-1)/n  integral ( cos t)^(n-2) dt + (cos t)^(n-1)sint/n where n = 4

 

the result is 3/4 integral cosu^2 du +  (cosu^3 * sin u)/4

 

The same reduction formula is used (recursively) to integrate cosu^2 in this result. It also shall be used to

integrate the first integral in this part of the mess.

 

Going back to location ALPHA earmarked above, the integral (9-x^2)^(1/2) is done

with the substitution x = 3 sin u as in the previous step.

The result is 9 integral cos u^2 which is used with the same reduction formula in the previous step.

 

Now it is time to return to location BETA and integral the first integral...

 

integral x^3 (9-x^2)^1/2 dx

 

Let U = 9 - x^2 ---> dx = -1/(2x) du

 

The integral becomes 1/2 integral u^(3/2) - 9 sqrt(u) du

 

This is easily integrated by splitting / linearity using the power rule:

 

This first part integrates to (9-x^2)^(5/2)/5 - 3(9-x^2)^(3/2)

 

The final (indefinite) integral is:

 

  [sqrt(9-x^2)*( 32x^4 - 190x^3-96x^2+855x-1728) - 7695*arcsin(x/3) ]/40

 

plugging in x=3 and x=-3 and subtracting the former from the latter,

 

the software says the result is 1539*pi/4 which is approximately 1208.73

 

 References:

http://www.wolframalpha.com/widget/widgetPopup.jsp?p=v&id=b41bfa792bb2b6b327cb7db6a92fa4e2&title=Triple%20Integral%20Calculator&theme=blue&i0=8xyz&i1=dzdxdy&i2=1&i3=2&i4=2&i5=3&i6=0&i7=1&podSelect=&includepodid=Input&showAssumptions=1&showWarnings=1

 

https://www.integral-calculator.com/