Calc 3 - evaluate the triple integral?
∫∫∫B ze^(x+y) dV = ∫(x = 0 to 3) ∫(y = 0 to 3) ∫(z = 0 to 1) ze^x e^y dz dy dx = ∫(x = 0 to 3) e^x dx * ∫(y = 0 to 3) e^y dy * ∫(z = 0 to 1) z dz = (e^3 - 1) * (e^3 - 1) * 1/2 = (1/2)(e^3 - 1)^2. I hope this helps!
Triple Integrals in Calc 3?
1) Let E be the region contained between the paraboloid x = y^2 + z^2 and the paraboloid x = -4 + 2y^2 +2z^2. When finding a triple integral of a function f over the region E, which variable should you integrate with respect to first? 2) Let E be the region contained between the cone y = radical(x^2 + z^2) and the plane y = 9. When finding a triple integral over the region E, which variable should you integrate with respect to first? 3) Suppose you are using a triple integral to find the volume of the region contained between the paraboloids z = x^2 + y^2 and z = 8 - x^2 - y^2. What is a valid setup for the triple integral?
Calc 3 Triple Integral Help?
z = sqrt(144 - (y^2)) set it to zero as z = 0 0 = sqrt(144 - (y^2)) y = -12 and 12 but keep the positive one only as your question mentions first octant as 12 = 4x as x = 3 and other x = 0 int_0^3 int_(4x)^12 int_0^sqrt(144 - (y^2)) z dz dy dx = 648
Calc 3 triple integral question?
∫∫∫ f(x,y,z) dV = ∫∫∫ r (2 cos(θ) + 3 sin(θ)) r dr dθ dz {(r,θ,z) | 0 ≤ r ≤ √2 ; π/4 ≤ θ ≤ 3π/4 ; 0 ≤ z ≤ 3} = 3 ∫∫ r² (2 cos(θ) + 3 sin(θ)) dr dθ {(r,θ) | 0 ≤ r ≤ √2 ; π/4 ≤ θ ≤ 3π/4} = 2√2 ∫ (2 cos(θ) + 3 sin(θ)) dθ {(θ) | π/4 ≤ θ ≤ 3π/4} = 12 Answer: 12
How do you visualize the volume bounded and integrated over in a triple integral that needs coordinate transformations?
Let's take an example.Q. The figure shows the region of integration for the integral[math] \int_{0}^{1} \int_{\sqrt{x}}^{1} \int_{0}^{1-y} f(x,y,z)~ dz~ dy ~dx [/math] Rewrite this integral as an equivalent iterated integral in the five other orders. Question from Stewart Calculus: Early Transcendentals, 7th Edition - Chapter 15, 15.7 Exercises, Q33. A. Whenever you have a question which already includes a diagram ie. where you do not have to construct the figure from scratch and then transform it into different coordinates, half your work is done right there. In the above questions, your multiple integral should have the following 5 transformations: (dx dz dy), (dx dy dz), (dz dx dy), (dy dz dx), and (dy dx dz).Here's the basic idea - for example, if we have to find the limits when we transform the coordinates to (dx dz dy), we first compress the x-axis, then the z-axis, and then the y-axis. To compress the x-axis, move from +x to -x and note the limits from x = g(y,z) to x = h(y,z) of the 3D figure. This means the limits should be in terms of either axis barring the one you are doing the transformation for.For our case, we move from [math]x = y^2~to~x = 0[/math]. Note these limits down and compress the x-axis. You can also think of this as projection of the figure in 3D space, onto the planes of the respective axes. Now compress the z-axis; move from +z to -z and note the limits from z = w(y) to z = v(y). In this case, we move from [math] z = 1-y ~to ~z = 0 [/math]. Finally, compress the y-axis which goes from [math]y = 1 ~to~ y = 0[/math]. Your final integral after the transformation should be:[math] \int_{0}^{1} \int_{0}^{1-y} \int_{0}^{y^2} f(x,y,z)~ dx~ dz~ dy [/math] Try this for the rest 4 transformations. Crosscheck with the answers.Answer from Stewart Calculus: Early Transcendentals, 7th Edition.Addendum: When you do not have the figure drawn for you and have to deduce it from the clues given in the question -> Go the other way around. Start from the origin, expand the line on the x,y,z axis then expand the the plane on the (x,y), (y,z) or (x,z) plane and make the 3D figure by expanding it in 3D space. This is just one trick I found helpful for solving the questions in my Calc 3 course. Hope this helps!
Use cylindrical coordinates to evaluate the triple integral ∫ ∫∫E √{x2 + y2} dV, where E... Calc 3 Help!?
First, get a picture of the solid E (either in your mind's eye, or on paper). It is an upside down solid dome whose bottom is a disc of radius 2 in the xy-plane. The top surface of the dome is given by the graph of z=4-(x^2+y^2). Next, use the formulas x=r cos theta y=r sin theta z=z to describe E and the integrand in terms of r, theta, and z. You can think of cylindrical coordinates as polar coordinates with z "along for the ride". So you can use the usual polar coordinate "tricks" that come from the above formulas, such as x^2+y^2=r^2. Describe E: r varies from 0 to 2, and theta varies from 0 to 2pi. For a particular choice of r and theta, the corresponding z varies from 0 to 4-(x^2+y^2)=4-r^2. Describe the integrand: sqrt(x^2+y^2)=sqrt(r^2)=r. Put it all together in an iterated integral--don't forget to throw in the extra r, just like with polar coordinates! int(0 to 2pi) int(0 to 2) int(0 to 4-r^2) r r dz dr d(theta). I think that you are most interested in the set-up, so I'll stop here. Hope that this reaches you in time and is helpful.
Using polar coordinates, evaluate the integral? Calc 2 help!!?
Note that the ring-shaped region can be described as 4 ≤ r ≤ 7 and 0 ≤ ≤ 2π. So, ∫∫R sin(x²+y²) dA = ∫(r = 4 to 7) ∫(θ = 0 to 2π) sin(r²) * (r dθ dr) = ∫(r = 4 to 7) 2πr sin(r²) dr = -π cos(r²) {for r = 4 to 7} = π (cos 16 - cos 49). I hope this helps!
Calc 3, Use cylindrical coordinates to Evaluate the triple integral over the region E?
Use cylindrical coordinates to evaluate. Triple integral over the region E of (x + y + z) dV where E is the solid in the first octant that lies under the paraboloid z = 9 − x^2 − y^2.
Calc 3 - polar integrals - please help!?
The annular region implies that r is in [1, 3] with θ in [0, 2π]. So, ∫∫D √(25 - x^2 - y^2) dA = ∫(θ = 0 to 2π) ∫(r = 1 to 3) √(25 - r^2) * (r dr dθ), converting to polar coordinates = ∫(θ = 0 to 2π) dθ * ∫(r = 1 to 3) r(25 - r^2)^(1/2) dr = 2π * [(2/3)(-1/2)(25 - r^2)^(3/2) {for r = 1 to 3}] = (2π/3) [24^(3/2) - 64] = (2π/3) (48√6 - 64). I hope this helps!
Triple Integral Calc 3 Question! Find the volume V of the pyramid with base in the plane z = 1 and sides....?
The bounds for z are immediately found to be z = 1 to z = 61 - 6x - y. Projecting the region onto the xy-plane yields a region bounded by y = 0, y = x + 4, and 6x + y + 1= 61 ==> y = 0, y = x + 4, and y = -6x + 60 ==> x = y - 4 to x = 10 - y/6 for y in [0, 12] (by finding where the slanted lines intersect). So, the volume ∫∫∫ 1 dV equals ∫(y = 0 to 12) ∫(x = y - 4 to 10 - y/6) ∫(z = 1 to 61 - 6x - y) dz dx dy. You can take it from here... I hope this helps!