Find the volume of the solid whose base is the region in the xy-plane bounded by the given curves and whose cross-sections perpendicular to the x-axis are (a) squares, (b) semicircles, and (c) equilateral triangles.
for y=x^2, x=0, and y=0
(a) integral (x^2)^2 from 0 to 2=32/5
(b) integral 1/2 pi (1/2x^2)^2 from 0 to 2 = 4pi/5
(c) integral 1/2x^2(1/2)(x^2)(sqrt(3)) from 0 to 2 = 8(sqrt(3))/5
for y=sqrt(x), x=0, x=16, and y=0
(a) integral (sqrt(x))^2 from 0 to 16= 128
(b) integral 1/2pi[(1/2)(sqrtx)]^2 from 0 to 16 = 16 pi
(c) integral 1/2 (sqrtx)(1/2)(sqrt(x))(sqrt(3))=32sqrt(3)
for y=8-x^2, y=x^2
(a) integral (8-2x)^2 from -2 to 2 = 192
(b) integral 1/2 pi [(8-2x^2)/2]^2 from -2 to 2 =256pi/15
(c) integral (8-2x^2)(sqrt3)(4-x^2) from -2 to 2 = 1024(sqrt3)/15
To find the volume of the solid whose base is the region bounded by the given curves and whose cross-sections are squares, semicircles, and equilateral triangles perpendicular to the x-axis, we can use the method of integration.
(a) For squares:
To find the volume of the solid when the cross-sections are squares, we need to integrate the area of each square over the interval. The area of each square can be found by squaring the length of each side.
In the first example, for the curve y = x^2, x = 0 to x = 2, the length of the side of each square is x^2. So, the volume can be found by integrating the area function (x^2)^2 from 0 to 2:
Volume = ∫[(x^2)^2]dx, over the interval [0, 2]
= ∫[x^4]dx, over the interval [0, 2]
= x^5/5 evaluated from 0 to 2
= [(2^5)/5 - (0^5)/5]
= 32/5
So, the volume of the solid when the cross-sections are squares is 32/5.
(b) For semicircles:
To find the volume of the solid when the cross-sections are semicircles, we need to integrate the area of each semicircle over the interval. The area of each semicircle can be found by taking half the area of the corresponding full circle.
In the first example, for the curve y = x^2, x = 0 to x = 2, the radius of each semicircle is (1/2)x^2. So, the volume can be found by integrating the area function (1/2)(π[(1/2)x^2]^2) from 0 to 2:
Volume = ∫[(1/2)(π[(1/2)x^2]^2)]dx, over the interval [0, 2]
= ∫[(1/2)(π)(1/4)(x^4)]dx, over the interval [0, 2]
= (π/8) ∫[x^4]dx, over the interval [0, 2]
= (π/8) (x^5/5) evaluated from 0 to 2
= (π/8) [(2^5)/5 - (0^5)/5]
= 4π/5
So, the volume of the solid when the cross-sections are semicircles is 4π/5.
(c) For equilateral triangles:
To find the volume of the solid when the cross-sections are equilateral triangles, we need to integrate the area of each triangle over the interval. The area of each equilateral triangle can be found by multiplying one-half the base length by the height, where the height is sqrt(3)/2 times the base length.
In the first example, for the curve y = x^2, x = 0 to x = 2, the base length of each triangle is (1/2)x^2, and the height is (sqrt(3)/2)(x^2). So, the volume can be found by integrating the area function (1/2)(x^2)(1/2)(x^2)(sqrt(3)) from 0 to 2:
Volume = ∫[(1/2)(x^2)(1/2)(x^2)(sqrt(3))]dx, over the interval [0, 2]
= (sqrt(3)/4) ∫[x^4]dx, over the interval [0, 2]
= (sqrt(3)/4) (x^5/5) evaluated from 0 to 2
= (sqrt(3)/4) [(2^5)/5 - (0^5)/5]
= 8(sqrt(3))/5
So, the volume of the solid when the cross-sections are equilateral triangles is 8(sqrt(3))/5.
Repeat the same process for the other given curves and their respective intervals to find the volumes for (a), (b), and (c) accordingly.