# Write a triple integral in cylindrical coordinates for the volume inside the cone

Spherical coordinates are especially convenient for working with solids bounded by these types of surfaces. Integration in Spherical Coordinates We now establish a triple integral in the spherical coordinate system, as we did before in the cylindrical coordinate system.

Sent by Atsuko Odoi on Thu, 21 Nov What is the maximum value of xyz subject to this constraint? Be sure to analyze carefully and completely any system of equations you solve.

## Integration - Finding volume of a cone using triple integral - Mathematics Stack Exchange

First, we use Lagrange Multipliers, by taking the gradient of the constraint, multiplying it by L which I will use for lambda hereand setting it equal to the gradient of the function we want to maximize.

This yields 3 equations, the constraint equation, and four unknowns. I will label each equation as: Then, we divide both sides by L. This will give me: If the first case is true, we have all three variables equal to zero. If the second case is true, we move on.

Now, since we have all the variables in terms of y, we can plug in for x and z into the constraint equation 4. When we plug in, we get 8. We plug this back into eq. We have eight candidates for max and min, but 4 of them will give the same value, and the other four will give the negative of that value.

Since these 8, and 0,0,0 are the only candidates for min and max, we know that the min is. And the max is. Sent by Alexander Pergament on Sun, 17 Nov First we notice that the field is conservative. But, since the beginning and final points are provided in the problem, we can simply rewrite this to be f 0,2 - f 0,0.

Note from the management: The parameterized integral can be done, and is really not too bad.

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The answer will again be 0. The computation above is much shorter, though! Sent by Greg Ryslik on Mon, 18 Nov Suppose R is the trapezoid with vertices 1,02,00,-2and 0, First, solve for x and y in terms of u and v.

So, the double integral becomes this iterated integral: So the inner integral is: Now use this to compute the outer integral.The Content - It's not just about batteries. Scroll down and see what treasures you can discover. Background.

\\begin{aligned}V&= 2\\iiint\\limits_T dxdydz= 2\\iiint\\limits_{T^{\\ast}}|J|\\,drd\\varphi dz= 2\\int\\limits_0^{2\\pi}d\\varphi \\int\\limits_0^{2\\sqrt a}r\\,dr \\int\\limits_0^{\\sqrt{2\\sqrt a r -r^2}}dz=\\\\ &=4\\pi\\int\\limits_0^{2\\sqrt a}r\\sqrt{2\\sqrt a r-r^2}\\,dr=\\ldots = 2\\pi^2a^{3/2}\\end{aligned}

We think of a battery today as a source of portable power, but it is no exaggeration to say that the battery is one of the most important inventions in the history of mankind. Section Triple Integrals in Cylindrical Coordinates In this section we want do take a look at triple integrals done completely in Cylindrical Coordinates.

Recall that cylindrical coordinates are really nothing more than an extension of polar coordinates into three dimensions. TRIPLE INTEGRALS IN SPHERICAL & CYLINDRICAL COORDINATES Triple Integrals in every Coordinate System feature a unique infinitesimal volume element.

In Rectangular Coordinates, the volume element, " dV " is a parallelopiped with sides: " dx ", " dy ", and " dz ". Accordingly, its volume is the product of its three sides, namely dV dx dy= ⋅ ⋅dz.

(2a): Triple integral in cylindrical coordinates r,theta,z Now the region D consists of the points (x,y,z) with x^2+y^2+z^2=sqrt(3)*r. Find the volume of this region.

Triple Integrals in Cylindrical Coordinates; Discussion; Triple Integrals in Spherical Coordinates; Summary. Triple Integrals in Cylindrical Coordinates. Cylindrical coordinates are obtained from Cartesian coordinates by replacing the x and y coordinates with polar coordinates r and theta and leaving the z coordinate unchanged.