Calculus! Find the volume.?

One limit is given, x = 1. The other one is where 7x² equals 0 which is x = 0. So the limits are set and now to find which function is greater on the interval. 7x² > 0 so 7x² is the greater function. Using the disc method is recommended here since the function to be rotated is in terms of x and the AoR is horizontal. Plugging all this in
V = π ∫[0,1] (7x²)² dx = 49π ∫[0,1] x⁴ dx

That's an elementary integral with simple limits so I believe you can take it from there.

If 2200 sq. centimeters of cloth is available to make a container with a sq. base and an open outstanding, discover the biggest available quantity of the container. quantity =____ cubic centimeters. quantity (V) includes length (L) * Width (W) * top (H). we would desire to maximise the quantity employing calculus. V = LWH = W²H (equation a million) all of us be attentive to: Ø A sq. base: potential L = W Ø container with front, back, 2 sides, and base section for the container is equivalent 2200 cm² and Base = W² front and back = HW + HW = 2HW 2 sides = HW + HW = 2HW comprehensive section = 2200 = W² + 2HW + 2HW = W² + 4HW So, 2200 = W² + 4HW (equation 2) Now take equation 2 and remedy for H and plug into equation a million 2200 = W² + 4HW [2200 – W²] / 4W = H simplify H = 550/W – (a million/4)W (equation 3) V = W² [550/W – (a million/4)W] simplify V(W) = 550W – (a million/4)W³ Now you will possibly desire to take by-fabricated from V(W), set it equivalent 0, and remedy for W V’(W) = (-3/4)W² + 550 = 0 W = ± (10?22)/3, W ? - (10?22)/3 on condition that W must be useful. W = 15.sixty 3 cm approximation H = 550/W – (a million/4)W = approximately 31.27 cm the biggest available quantity is W²*H = (15.sixty 3)²(31.27) = 7643.seventy 8 cm³

y = 7x^2

Use the Disk method:

dV = πy^2 dx = π (7x^2)^2 dx = π (49x^4) dx

V = π ∫ 49x^4 dx = π (49x^5/5) {from x = 0 to x = 1}

Evaluating from x = 0 to x = 2: π (49/5 - 0) = 49π/5