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Problem. 8 : Let E be the solid that lies within the cylinder x^2 + y^2 = 1, above the plane z = 0 and below the cone z^2 = 25x^2 + 25y^2. Express the following integral in cylindrical coordinates. ?_E (x^2 + y^2)^{3/2} dV = ?_0^{2pi} ?_0^1 ?_0^{5r} r^4 dz dr d? Problem. 9 : Set up a triple integral in cylindrical coordinates of the solid that lies within the cylinder x^2 + y^2 = 16 and the sphere x^2 + y^2 + z^2 = 49. V = ?_0^{2pi} ?_0^4 ?_{-sqrt(-r^2+49)}^{sqrt(-r^2+49)} r dz dr d? Problem. 10 : Set up a triple integral in cylindrical coordinates of the solid that lies inside the sphere x^2 + y^2 + z^2 = 162 and above the upper nappe of the cone z^2 = x^2 + y^2 in the first octant. V = ?_0^{1/2pi} ?_0^9 ?_r^{sqrt(-r^2+162)} r dz dr d?

          Problem. 8 : Let E be the solid that lies within the cylinder x^2 + y^2 = 1, above the plane z = 0 and below the cone z^2 = 25x^2 + 25y^2. Express the following integral in cylindrical coordinates.

?_E (x^2 + y^2)^{3/2} dV = ?_0^{2pi} ?_0^1 ?_0^{5r} r^4 dz dr d?

Problem. 9 : Set up a triple integral in cylindrical coordinates of the solid that lies within the cylinder x^2 + y^2 = 16 and the sphere x^2 + y^2 + z^2 = 49.

V = ?_0^{2pi} ?_0^4 ?_{-sqrt(-r^2+49)}^{sqrt(-r^2+49)} r dz dr d?

Problem. 10 : Set up a triple integral in cylindrical coordinates of the solid that lies inside the sphere x^2 + y^2 + z^2 = 162 and above the upper nappe of the cone z^2 = x^2 + y^2 in the first octant.

V = ?_0^{1/2pi} ?_0^9 ?_r^{sqrt(-r^2+162)} r dz dr d?

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Problem. 8 : Let E be the solid that lies within the cylinder x^2 + y^2 = 1, above the plane z = 0 and below the cone z^2 = 25x^2 + 25y^2. Express the following integral in cylindrical coordinates.

?E (x^2 + y^2)^3/2 dV = ?0^2pi ?0^1 ?0^5r r^4 dz dr d?

Problem. 9 : Set up a triple integral in cylindrical coordinates of the solid that lies within the cylinder x^2 + y^2 = 16 and the sphere x^2 + y^2 + z^2 = 49.

V = ?0^2pi ?0^4 ?-sqrt(-r^2+49)^sqrt(-r^2+49) r dz dr d?

Problem. 10 : Set up a triple integral in cylindrical coordinates of the solid that lies inside the sphere x^2 + y^2 + z^2 = 162 and above the upper nappe of the cone z^2 = x^2 + y^2 in the first octant.

V = ?0^1/2pi ?0^9 ?r^sqrt(-r^2+162) r dz dr d?

Added by Makayla G.

Calculus: Early Transcendentals

James Stewart 8th Edition

Instant Answer

Solved by Expert Sri K

05/25/2022

Step 1

Converting these equations to cylindrical coordinates, we get: \(z = r^2\) and \(2z = 25 - 25r^2\) Solving for z in the cone equation, we get \(z = \frac{25 - 25r^2}{2}\) Show more…

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Problem: Let E be the solid that lies within the cylinder x^2 + y^2 = 1, above the plane z = 0 and below the cone z^2 = 25x^2 + 25y^2. Express the following integral in cylindrical coordinates: ∭ (x^2 + y^2)^{3/2} dV = ∫_0^{2pi} ∫_0^1 ∫_0^{5r} r^4 dz dr dθ Problem: Set up a triple integral in cylindrical coordinates of the solid that lies within the cylinder x^2 + y^2 = 16 and the sphere x^2 + y^2 + z^2 = 49: V = ∫_0^{2pi} ∫_0^4 ∫_{-sqrt(-r^2+49)}^{sqrt(-r^2+49)} r dz dr dθ Problem: Set up a triple integral in cylindrical coordinates of the solid that lies inside the sphere x^2 + y^2 + z^2 = 162 and above the upper nappe of the cone z^2 = x^2 + y^2 in the first octant: V = ∫_0^{1/2pi} ∫_0^9 ∫_r^{sqrt(-r^2+162)} r dz dr dθ

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Transcript

00:01 Here for the first subpart, we have x is equal to r cost theta and y is equal to r sine theta.

00:12 Now, these are the conversion equations from cartesian to polar coordinates.

00:17 Also, we have z square is equal to 25 multiplied by x square plus y square.

00:23 Now observe that here 25 multiplied by value of x as r cost theta, so x square will be r square cosper theta plus similarly r squared sine square theta.

00:40 Now we know that sine square theta plus cos square theta is equal to 1.

00:45 So we are left with only r square.

00:48 So we have z square is equal to 25 r square.

00:53 So taking square roots, we have z is equal to 5r.

01:00 So we have the limits of z as from 0 to 5r.

01:05 Now we have the circle x square plus y square equal to 1.

01:14 Now similarly as about we have x square plus y square as r square.

01:20 So r square is equal to 1.

01:22 Now we have the limits of r as r is equal to 0 to r is equal to 1.

01:34 And also we have the limits of theta as theta is equal to 0 to 2 pi.

01:41 Now we have the limits z is equal to 0 to 5r.

01:46 Also r is equal to 0 to r is equal to 1 and the theta is equal to 0 to 2 pi.

01:56 So further to the triple integral, we have triple integral of x square plus y square, whole bracket raised to 3 divided by 2 dv...

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