# Volume Of A Solid Calculus

Volume of a Solid- Calculus?

The base of a certain solid is the triangle with vertices (0,0), (3,0), and (0,6). Cross-sections perpendicular to the y-axis are isosceles triangles with height equal to the base.

Use the formula V=∫baA(y)dy to find the volume of the solid.

The lower limit of integration is a =

The upper limit of integration is b =

The base of the triangular cross-section is the following function of y:

The area of the triangular cross-section is A(y)=

Thus the volume of the solid is V =

I know the lower limit is 0 and the upper limit is 6, but unsure of the next steps...

Calculus question - Volume of a solid?

Okedoke.

First of all, the radius of the circle forming the base is 7. No argument, right?

Therefore, a cross-section perpendicular to the x-axis at x=0 would yield a triangle with base of size 2r = 14 across. Therefore the cone is 14 units high.

The volume of a right cone = 1/3 * pi * r^2 * h = 1/3 * pi * 49 * 14 = 686pi/3 cubic units.

The area of a vertical cross section at x = 5? Well, what we need to do is find the size of the "base" of the triangle, which will also be its height, then we'll prettymuch be finished.

For simplicity, imagine just the first quadrant of the x-y plane with the circular curve described by the function y = √(49 - x^2)

When x = 5, y = 2√6.

THAT is the length from the x-axis to one side of the circle (half the base of the triangle in the cross section). Cutting all the way across the circle will be 2 times this, or 4√6, as the base of the triangle. This will also be the height.

Therefore, Area = 1/2 * (4√6) * (4√6) = 48 square units.

Volume of a solid-Calculus?

Consider the solid S described below.

The base of S is the region enclosed by the parabola y = 10 - 10x2 and the x-axis. Cross-sections perpendicular to the x-axis are isosceles triangles with height equal to the base.

Find the volume V of this solid.

Volume of a solid--Calculus?

area of equilateral triangle... √(3) s²/ 4 , s - side

thus for your solid...

V = ∫ (area of triangle) dy

the border is simply... x = (1 - y)

while x = side of such an equilateral triangle...

thus

V = ∫ √3 (1 - y)² / 4 dy .. {from 0 to 1}

= √3 /4 ∫ (1 - 2y + y²) dy

= √3 /4 (y - y² + y³/3)

= √3 /4 (1 - 1 + 1/3)

= √3 / 12

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CALCULUS HELP!!! Find the volume of a solid using integrals.?

I'll do #1

First graph the functions and find the region. Rotate this region about the x-axis to get yourself a good picture. Then you draw an arbitrary cross section of the region which will be a circle.

The area of a circle is pir^2 so to find the volume you need to do the integral from -1 to 2 (the points at which the functions intersect) of pi((x+3)^2) -pi((x^2)+1)^2. (the functions are the radii of the circles)

pi is a constant which you can put in front of the integral. so you are left with pi times the integral of (x+3)^2 - (x^2+1)^2

After you find the integrals multiply by pi and simplify you get -pi/5(x^5)-pi/3(x^3)+3pi(x^2)+8pi(x)

[-pi/5((2)^5)-pi/3((2)^3)+3pi((2)^2)+8...

Your numerical answer will be: 117pi/5 or 73.513

Could you explain simply what is solid of revolution calculus?

Let's say you have a curve y=f(x). There is area below the curve. Now imagine this curve (and area) rotates by 360 degrees about the x-axis. The curve will start sweeping through 3d space and the space which it sweeps through becomes a volume. After a full rotation you will get a solid body that encloses certain volume. This is called solid of revolution because you get a solid object whose volume is occupied.

Find the volume of a solid.. Calculus help please!?

This is an Area of a Cross-Section problem.

Area of a square = b²

Since the base, in this case, is perpendicular to the x-axis, you know that the base is basically equal to y, since y is the perpendicular distance from the x-axis.

Therefore, Area of a square = y²

What's y? y=sin^-1(x).

Therefore, Area of a square = (sin^-1(x))² or sin^-2(x).

Now, take the integral from 0 to 1 of the area of the square.

You'll need your calculator and fnInt.

fnInt((sin^-1(x))²),x,0,1) = 0.467

Finding the volume of a solid using calculus?

Consider the solid S described below.

The base of S is a circular disk with radius r. Parallel cross-sections perpendicular to the base are squares.

Find the volume V of this solid.

So my thought is take the area and take the integral of that from 0 to r. Is this correct? This is what I have so far.

V(x) = integral from 0 to r(pi*r^2 dx) = pi*r^3