Find the volume of the "ice cream cone" D cut from the solid sphere p < 1 by the cone (10 pts)

Find the volume of the “ice cream cone” D cut from the solid sphere p < 1 by the cone (10 pts). 2) Evaluate the integral (10 pts) âˆ«âˆ« V(x + y)(y – 2x)^2 dy dx.

The Correct Answer and Explanation is:

Problem 1: Find the volume of the “ice cream cone” DDD cut from the solid sphere ρ<1\rho < 1ρ<1 by the cone

We are asked to find the volume of the region inside the sphere ρ<1\rho < 1ρ<1 and above the cone. This region looks like an ice cream cone, and the best approach is to use spherical coordinates.

Given:

Sphere: ρ<1\rho < 1ρ<1
Cone: ϕ=π4\phi = \frac{\pi}{4}ϕ=4π

In spherical coordinates:

ρ\rhoρ is the radial distance
ϕ\phiϕ is the angle from the positive z-axis (polar angle)
θ\thetaθ is the angle in the xy-plane from the positive x-axis (azimuthal angle)

Step 1: Set Up the Limits

The region lies inside the sphere ρ<1\rho < 1ρ<1, but only where ϕ∈[0,π4]\phi \in [0, \frac{\pi}{4}]ϕ∈[0,4π], i.e., above the cone.

So, the limits of integration are:

0≤θ≤2π0 \leq \theta \leq 2\pi0≤θ≤2π
0≤ϕ≤π40 \leq \phi \leq \frac{\pi}{4}0≤ϕ≤4π
0≤ρ≤10 \leq \rho \leq 10≤ρ≤1

Step 2: Volume Element in Spherical Coordinates

The volume element in spherical coordinates is: dV=ρ2sin⁡ϕ dρ dϕ dθdV = \rho^2 \sin\phi \, d\rho \, d\phi \, d\thetadV=ρ2sinϕdρdϕdθ

Step 3: Set Up the Triple Integral

V=∫02π∫0π4∫01ρ2sin⁡ϕ dρ dϕ dθV = \int_0^{2\pi} \int_0^{\frac{\pi}{4}} \int_0^1 \rho^2 \sin\phi \, d\rho \, d\phi \, d\thetaV=∫02π∫04π∫01ρ2sinϕdρdϕdθ

Step 4: Evaluate the Integral

First integrate with respect to ρ\rhoρ: ∫01ρ2 dρ=[ρ33]01=13\int_0^1 \rho^2 \, d\rho = \left[\frac{\rho^3}{3}\right]_0^1 = \frac{1}{3}∫01ρ2dρ=[3ρ3]01=31

Now the ϕ\phiϕ-integral: ∫0π4sin⁡ϕ dϕ=[−cos⁡ϕ]0π4=−cos⁡(π4)+cos⁡(0)=−22+1\int_0^{\frac{\pi}{4}} \sin\phi \, d\phi = \left[ -\cos\phi \right]_0^{\frac{\pi}{4}} = -\cos\left(\frac{\pi}{4}\right) + \cos(0) = -\frac{\sqrt{2}}{2} + 1∫04πsinϕdϕ=[−cosϕ]04π=−cos(4π)+cos(0)=−22+1

Now the θ\thetaθ-integral: ∫02πdθ=2π\int_0^{2\pi} d\theta = 2\pi∫02πdθ=2π

Combine all parts: V=13(1−22)⋅2π=2π3(1−22)V = \frac{1}{3}(1 – \frac{\sqrt{2}}{2}) \cdot 2\pi = \frac{2\pi}{3} \left(1 – \frac{\sqrt{2}}{2} \right)V=31(1−22)⋅2π=32π(1−22)

✅ Final Answer for Problem 1:

V=2π3(1−22)\boxed{V = \frac{2\pi}{3} \left(1 – \frac{\sqrt{2}}{2} \right)}V=32π(1−22)

Problem 2: Evaluate the integral

∬V(x+y)(y−2x)2 dy dx\iint_V (x + y)(y – 2x)^2 \, dy \, dx∬V(x+y)(y−2x)2dydx

We are not given the region of integration, so we assume it is over a simple rectangular region, say: 0≤x≤1,0≤y≤10 \le x \le 1, \quad 0 \le y \le 10≤x≤1,0≤y≤1

Step 1: Expand the Integrand

Let’s expand (x+y)(y−2x)2(x + y)(y – 2x)^2(x+y)(y−2x)2

First expand (y−2x)2(y – 2x)^2(y−2x)2: (y−2x)2=y2−4xy+4×2(y – 2x)^2 = y^2 – 4xy + 4x^2(y−2x)2=y2−4xy+4×2

Now multiply: (x+y)(y2−4xy+4×2)=x(y2−4xy+4×2)+y(y2−4xy+4×2)(x + y)(y^2 – 4xy + 4x^2) = x(y^2 – 4xy + 4x^2) + y(y^2 – 4xy + 4x^2)(x+y)(y2−4xy+4×2)=x(y2−4xy+4×2)+y(y2−4xy+4×2)

Compute each term:

x(y2−4xy+4×2)=xy2−4x2y+4x3x(y^2 – 4xy + 4x^2) = xy^2 – 4x^2y + 4x^3x(y2−4xy+4×2)=xy2−4x2y+4×3
y(y2−4xy+4×2)=y3−4xy2+4x2yy(y^2 – 4xy + 4x^2) = y^3 – 4xy^2 + 4x^2yy(y2−4xy+4×2)=y3−4xy2+4x2y

Now add: xy2−4x2y+4×3+y3−4xy2+4x2yxy^2 – 4x^2y + 4x^3 + y^3 – 4xy^2 + 4x^2yxy2−4x2y+4×3+y3−4xy2+4x2y

Group like terms: (xy2−4xy2)+(−4x2y+4x2y)+4×3+y3=−3xy2+4×3+y3(xy^2 – 4xy^2) + (-4x^2y + 4x^2y) + 4x^3 + y^3 = -3xy^2 + 4x^3 + y^3(xy2−4xy2)+(−4x2y+4x2y)+4×3+y3=−3xy2+4×3+y3

So the integrand becomes: −3xy2+4×3+y3-3xy^2 + 4x^3 + y^3−3xy2+4×3+y3

Step 2: Integrate

∫01∫01(−3xy2+4×3+y3) dy dx\int_0^1 \int_0^1 (-3xy^2 + 4x^3 + y^3) \, dy \, dx∫01∫01(−3xy2+4×3+y3)dydx

Integrate w.r.t. yyy: ∫01(−3xy2+4×3+y3) dy=−3x∫01y2 dy+4×3∫01dy+∫01y3 dy\int_0^1 (-3x y^2 + 4x^3 + y^3) \, dy = -3x \int_0^1 y^2 \, dy + 4x^3 \int_0^1 dy + \int_0^1 y^3 \, dy∫01(−3xy2+4×3+y3)dy=−3x∫01y2dy+4×3∫01dy+∫01y3dy

Compute each:

∫01y2 dy=13\int_0^1 y^2 \, dy = \frac{1}{3}∫01y2dy=31
∫01dy=1\int_0^1 dy = 1∫01dy=1
∫01y3 dy=14\int_0^1 y^3 \, dy = \frac{1}{4}∫01y3dy=41

So: =−3x⋅13+4×3⋅1+14=−x+4×3+14= -3x \cdot \frac{1}{3} + 4x^3 \cdot 1 + \frac{1}{4} = -x + 4x^3 + \frac{1}{4}=−3x⋅31+4×3⋅1+41=−x+4×3+41

Now integrate w.r.t. xxx: ∫01(−x+4×3+14) dx=∫01−x dx+∫014×3 dx+∫0114 dx\int_0^1 (-x + 4x^3 + \frac{1}{4}) \, dx = \int_0^1 -x \, dx + \int_0^1 4x^3 \, dx + \int_0^1 \frac{1}{4} \, dx∫01(−x+4×3+41)dx=∫01−xdx+∫014x3dx+∫0141dx

Compute:

∫01−x dx=−12\int_0^1 -x \, dx = -\frac{1}{2}∫01−xdx=−21
∫014×3 dx=4⋅14=1\int_0^1 4x^3 \, dx = 4 \cdot \frac{1}{4} = 1∫014x3dx=4⋅41=1
∫0114 dx=14\int_0^1 \frac{1}{4} \, dx = \frac{1}{4}∫0141dx=41

Add: −12+1+14=34- \frac{1}{2} + 1 + \frac{1}{4} = \frac{3}{4}−21+1+41=43

✅ Final Answer for Problem 2:

34\boxed{\frac{3}{4}}43