Let be the volume of a pyramid of height 15 whose base is a square of side 7. Us

Question

Let be the volume of a pyramid of height 15 whose base is a square of side 7.

Use similar triangles to find the area of the horizontal cross section at a height .

Calculate by integrating the cross-sectional area.

Let V be the volume of a pyramid of height 15 whose base is a square of side 7. Use similar triangles to find the area of the horizontal cross section at a height y. A = Calculate V by integrating the cross-sectional area. V = Find the volume of liquid needed to fill a sphere of radius R to height R/2. Find the volume of the solid whose base is the circle x2 + y2 = 36 and the cross sections perpendicular to the x-axis are triangles whose height and base are equal. Find the area of the vertical cross section A at the level x = 5. A= V= Find the volume of the solid whose base is the semicircle y = 4 - x2 where -2 x 2, and the cross sections perpendicular to the x-axis are squares. Find the volume of the solid whose base is the region enclosed by y = x2 and y = 3, and the cross sections perpendicular to the y-axis are squares. The average value of f(x) = 2x3 - 16x2 on the interval [-5, 8] is

Explanation / Answer

1)

Similar triangles have similar ratios, so:
4/r=10/(10-y)
4-.4y=r
area of large triangle = 1/2ry=1/2(4)(10)=20
20-1/2ry= A large triangle minus small triangle= A =area
20-1/2(4-.4y)y=A substitute r
.2y^2-2y+20=A

integrate this, gives

y((.066667y-1)y+20)=volume
solve for y=10, gives 166.67

2)

For a sphere with radius = r

Let h be the height of liquid in the sphere.

If we subdivide the height into thin segments by dh, let x be the radius at height h.

So when h is 0, then x is 0 (empty sphere)
and when h is r, then x is r (1/2 full sphere {known to be 2/3 PH r^3}

We know that the circumfrential area for any little dh, is CA = PI * x^2

We need the relationship between x and h.

As h increases, the remaining distance from the center of the sphere to the top of the liquid is r-h.

The length of the line from the center of the sphere to the edge of the liquid is r

So we can find x using x^2 + (r-h)^2 = r^2
x^2 = r^2 - {r^2 -hr -hr + h^2}
x^2 = r^2 - {r^2 -2hr + h^2}
x^2 = r^2 - r^2 + 2hr - h^2
x^2 = + 2hr - h^2
x = {2hr -h^2}^(1/2)

3)

A = 1/2 (2 ?(36 - x

Related Questions

Navigate

• Browse (All)
• Browse L
• Subjects

---

---

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