NUMBER 10 Use spherical coordinates to evaluate the following triple integral; i

Question

NUMBER 10

Use spherical coordinates to evaluate the following triple integral; integral_-2^2 integral_0^Squareroot 4-y^2 integral_- Squareroot 4-x^4-y^2^Squareroot 4-x^4-y^2 y^2 Squareroot x^2 + y^2 + z^2 dz dx dy Use the transformation u = x -y, v = x + y to evaluate the double integral integral integral_R x-y/x+y dA where R is the square with vertices (0.2), (1, 1), (2, 2), and (1, 3). Find the surface area of the part of the surface f(x, y) = x^2 + that lies above the triangle with vertices (0, 0), (1, 0), and (0.2). Rewrite the triple integral integral--1^1 integral_x^2^1 integral_01-y f(x, y, z)dz, dy dx as an iterated triple integral with the order of integration as dx dy dz. Find the volume of the solid region that lies above the paraboloid z = x^2 + y^2 and below the half-cone z = Squareroot x^2 + y2. Use cylindrical coordinates to evaluate integral integral_E integral z dV, where E is the region enclosed by the paraboloid z = x^2 + y^2 and the plane z = 4. Evaluate the integral integral integral_R (x + y)e^x^2-y^2 dA y making an appropriate change of variables, where R is the rectangle enclosed by the lines x-y = 0, x-y = 2, x + y = 0, and x + y = 3. Find the average value of f(x, y) = e^y Squareroot x + e^y over the rectangle R = [0, 4] x [0, 1]. Let E be the pyramid with vertices A(0.0, 0), B(5.0.0), C(0, 1, 0). and D(0.0, 40). Set up the integral integral integral_E integral f(x, y, z)dV as a triple iterated integral with the order of integration given by dx dy dz.

Explanation / Answer

fx = 1
(fx)^2 = 1
fy = 2y
(fy)^2 = 4y^2

We have to Use the formula that consist of double integral and dA .

But there is this (1+ (fy)^2 + (fx)^2 ) ^0.5

We will get

(2+4y^2)^0.5

Setting up double integral, we will be integrating with respect to x first then y because you have an expression wich consist of y.
outermost should have the bounds 0 to 1 (this will be for dy)
innermost bounds should be from 0 to y

Then factoring out a square root of 2 to get and then integrating with respect to x to get

(square root of 2) times integral from 0 to 1 for the expression y(1+2y^2) dy

Use u-sub of u = 1+2y^2, du = 4y dy, du/4 = y dy

So we are getting root 2 divided by 4 times integral from 1 to 3 (u^0.5) du

Integrating and substituting the bounds, you get root 2 divided by 6 and this is multiplied by 3^(1.5) - 1

Related Questions

Navigate

• Browse (All)
• Browse N
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.