In the following problem 1 you use the concept of \"volume of a solid between tw

Question

In the following problem 1 you use the concept of "volume of a solid between two surfaces" that I introduced in class, (you need to know which is the upper and which is the lower surface that bound the solid, i.e. you need to draw yourself a picture of it, in addition, you also need to know the 2-dimensional region D that is the projection of the solid into the xy-plane, because that is the region D over which you integrate, you find that usually by finding the equation in x and y of the curve of intersection of the two surfaces (we'll do examples in class)) Use polar coordinates to find the volume of the given solid that a. lies below the paraboloid z = 18 - 2x^2 - 2y^2 and above the xy-plane b. lies above the cone z = Squareroot x^2 + y^2 and below the sphere x^2 + y^2 + z^2 = 1 c. is bounded by the paraboloid z - 1 + 2x^2 + 2y^2 and the plane z = 7 d. lies inside both the cylinder x^2 + y^2 = 4 and the ellipsoid 4x^2 + 4y^2 + z^2 = 64 Evaluate the iterated integral by converting to polar coordinates a. integral_0^a integral_-Squareroot a^2 - y^2^0 x^2 y dx dy b. integral_0^1 integral_y^Squareroot 2 - y^2 x + y dx dy

Explanation / Answer

Solution:2(a)

the region of integration is x = -(a^2 y^2) to x = 0 with y in [0, a].
==> Inside x^2 + y^2 = a^2 in the first quadrant under x = 0
==> Inside r = a with in [0, /4] using polar coordinates.

So, (y = 0 to a) (x = -(a^2 y^2) to 0) x^2 y dx dy
= ( = 0 to /4) (r = 0 to a) (r cos )^2 (r sin ) * (r dr d), via polar coordinates
= ( = 0 to /4) sin cos^2 d * (r = 0 to a) r^4 dr
= -(1/3) cos^3() {for = 0 to /4} * (1/5)r^5 {for r = 0 to a}
= -(1/3){(1/2)^3 - 1} * (1/5)a^5
= -(1/15){1/22 - 1}a^5

Solution:2(b)

The region of integration is x = y to x = (2y^2) with y in [0, 1].
==> Inside x^2 + y^2 = 2 in the first quadrant under y = x.
==> Inside r = 2 with in [0, /4] using polar coordinates.

So, (y = 0 to 1) (x = y to (2 - y^2)) (x+y) dx dy
= ( = 0 to /4) (r = 0 to 2) (r cos + r sin ) * (r dr d), via polar coordinates
= ( = 0 to /4) d * {(1/3)r^3 cos + (1/3)r^3 sin } {for r = 0 to 2}
= ( = 0 to /4) {(22/3) cos + (22/3) sin } d
= (22/3) {sin() - cos()} {for = 0 to /4}
= (22/3) [{1/2 - 1/2} - {0 - 1}]
= (22/3)

Related Questions

Navigate

• Browse (All)
• Browse I
• Subjects

---

---

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