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Since we have x in terms of y and are rotating about a vertical axis, use washers.

Each washer will have outer radius R = x = √5y²

and inner radius r = 0, so

V = ∫[a,b] π(R² - r²) dy = π∫[-1,1] (√5y²)² dy = 5π∫[-1,1] y^4 dy

V = πy^5 |[-1,1] = π(1 - -1) = 2π

1. Well, that's just the volume of a sphere of radius 3, V = (4π/3)r³ = 36π.

If you MUST use calculus, then use the disk (washer) method.

Each disk has radius R = y = √(9 - x²), so

V = ∫[-3,3] π(9 - x²) dx = π(9x - (1/3)x³) |[-3,3] = 2π(27 - 9) = 36π

2. Again, disks. R = secx, so

V = π∫[-π/4,π/4] sec²x dx = πtanx |[-π/4,π/4] = π(1 - -1) = 2π

3. Oy, enough with the disks already. R = x = 1 - y².

V = π∫[-1,1] (1 - y²)² dy = π∫[-1,1] (1 - 2y² + y^4) dy = π(y - (2/3)y³ + (1/5)y^5) |[-1,1]

V = 2π(1 - 2/3 + 1/5) = (2π/15)(15 - 10 + 3) = 16π/15

11. Each square will have a side b = y = √(16 - x²)

and therefore has area A = b² = 16 - x²

Add up all these little "dx" squares.

V = ∫[-4,4] (16 - x²) dx = (16x - (1/3)x³) |[-4,4] = 2(64 - 64/3) = 256/3 ≈ 85

x = √(5)y²

This is an even function.

We can rewrite it with x as the subject:

y = ±√(x)/5^(1/4)

Since it's an even function, we can use just the top half to find the volume, then multiply it by 2.

The bounds then change to y[0,1], which maps to x[0,√5]

Shell Method:

V = 2 * [π ∫ x f(x) dx]

V = 2π/5^(1/4) ∫[0,√5] x^(3/2) dx

... integrate that

we are able to attempt this by ability of disks technique or cylindrical shells. i am going to do exactly them both, so that you'll merely pick which technique you want. i'm assumingthat you are able to imagine the best generated after revolution. DISKS technique: Disks technique require that the elements be perpendicular to the axis of rotation. because that your axis of rotation is the vertical line x = 2, then we will be making use of horizontal factors of thickness dy. If it were vertical factors, we'd want to be making use of dx. because that we will be making use of dy, we opt to go back up with the obstacles as y-coordinates, and the expressions in words of y, or a minimum of isolating an x. because it really is bounded from the x-axis, of route a million sure might want to be 0 for y. Now, for the better sure, the curve y = x^3 intersects the line x = 2 at (2,8) {merely plug in 2 for x in y = x^3, and also you word why y should be 8}. subsequently, our bounds variety from 0 as a lot as 8. Now, for the expressions to be in words of y, or remoted x's, we in effortless words want 2. x = 2 {remoted x}; and y = x^3 -> x = cbrt(y) {expression in words of y} Now, installation the imperative, we've V = pi * int (from 0 to eight) {(2 - cbrt(y))^2} dy V = pi * int (from 0 to eight) {4 - 4cbrt(y) + cbrt(y^2)} dy V = pi * [4y - 3y^(4/3) + 3/5 y^(5/3)] |from 0 to eight V = pi * [4(8 - 0) - 3(8^(4/3) - 0^(4/3)) + 3/5 (8^(5/3) - 0^(5/3))] V = pi * [32 - 40 8 + ninety six/5] V = 16/5 pi all top! we've our volume by ability of disks technique. we are able to now use cylindrical shells to ascertain, and rather technique. CYLINDRICAL SHELLS technique: Cylindrical shells require factors that are parallel to the axis of rotation. So for this one, we want vertical factors. So we want dx factors. making use of dx, we want bounds for x, and all expressions are in words of x or remoted y. For the obstacles alongside x, we've the beginning, so one x-sure might want to be 0, and because that it is going to the line x = 2, yet another sure might want to be at 2. We already have the expression y = x^3, so we are reliable to set it up. V = 2pi * int (from 0 to 2) {(2 - x) * x^3} dx V = 2pi * int (from 0 to 2) {2x^3 - x^4} dx V = 2pi * [a million/2 x^4 - a million/5 x^5] |from 0 to 2 V = 2pi * [a million/2 (2^4 - 0^4) - a million/5 (2^5 - 0.5)] V = 2pi * [8 - 32/5] V = 16/5 pi Which consents with the reply previously. i concept that you comprehend the recommendations in the back of disks technique and cylindrical shells technique. desire this helps!!!!!