Triple Integrals Bc4De7

1. **Problem Statement:** Evaluate the triple integrals in part (a), find the volume in part (b), and convert the integral in part (c) to cylindrical coordinates. --- 2. **Part (a) i:** Evaluate $$\int_0^2 \int_0^4 \int_x^2 yze^{x^3} \, dx \, dy \, dz$$. - The order of integration is $$dx$$, then $$dy$$, then $$dz$$. - Since the integrand is $$yze^{x^3}$$, and the inner integral is with respect to $$x$$ from $$x$$ to $$2$$, note the lower limit depends on $$x$$ itself, which is unusual. This likely means the inner integral is $$\int_x^2 yze^{x^3} \, dx$$ where the variable of integration is $$x$$, but the lower limit is $$x$$, which is the variable of integration itself. This is ambiguous; assuming the limits are constants or a typo, we interpret the integral as $$\int_0^2 \int_0^4 \int_x^2 yze^{x^3} \, dx \, dy \, dz$$ with inner integral $$x$$ from $$x$$ to $$2$$. - Since the inner integral variable and limit are the same symbol, we rename the inner variable to $$t$$ to avoid confusion: $$\int_0^2 \int_0^4 \int_t= x \text{ to } 2 yze^{t^3} \, dt \, dy \, dz$$. - The inner integral: $$\int_x^2 yze^{t^3} \, dt = y z \int_x^2 e^{t^3} \, dt$$. - The integral $$\int e^{t^3} dt$$ has no elementary antiderivative, so we keep it as is. - The outer integrals over $$y$$ and $$z$$ are straightforward. - Since $$y$$ and $$z$$ are independent of $$t$$, the integral becomes: $$\int_0^2 \int_0^4 y z \left( \int_x^2 e^{t^3} dt \right) dy dz$$. - Integrate over $$y$$ from $$0$$ to $$4$$: $$\int_0^4 y dy = \frac{4^2}{2} = 8$$. - Integrate over $$z$$ from $$0$$ to $$2$$: $$\int_0^2 z dz = \frac{2^2}{2} = 2$$. - So the integral reduces to: $$8 \times 2 \times \int_0^2 e^{t^3} dt = 16 \int_0^2 e^{t^3} dt$$. - Since $$\int_0^2 e^{t^3} dt$$ has no elementary form, the answer is: $$16 \int_0^2 e^{t^3} dt$$. --- 3. **Part (a) ii:** Evaluate $$\int_0^1 \int_0^{\sqrt{1-x^2}} \int_0^{\sqrt{1-x^2-y^2}} \frac{dz \, dy \, dx}{1 + x^2 + y^2 + z^2}$$. - The region of integration is the upper quarter of the unit ball in the first octant. - Use spherical coordinates: $$x = \rho \sin \phi \cos \theta$$, $$y = \rho \sin \phi \sin \theta$$, $$z = \rho \cos \phi$$, where $$\rho \in [0,1]$$, $$\phi \in [0, \frac{\pi}{2}]$$, $$\theta \in [0, \frac{\pi}{2}]$$. - The volume element is: $$dV = \rho^2 \sin \phi \, d\rho \, d\phi \, d\theta$$. - The integrand becomes: $$\frac{1}{1 + \rho^2}$$. - The integral becomes: $$\int_0^{\pi/2} \int_0^{\pi/2} \int_0^1 \frac{\rho^2 \sin \phi}{1 + \rho^2} \, d\rho \, d\phi \, d\theta$$. - Separate integrals: $$\left( \int_0^{\pi/2} \sin \phi \, d\phi \right) \left( \int_0^{\pi/2} d\theta \right) \left( \int_0^1 \frac{\rho^2}{1 + \rho^2} d\rho \right)$$. - Compute each: $$\int_0^{\pi/2} \sin \phi \, d\phi = 1$$, $$\int_0^{\pi/2} d\theta = \frac{\pi}{2}$$, $$\int_0^1 \frac{\rho^2}{1 + \rho^2} d\rho = \int_0^1 \left(1 - \frac{1}{1 + \rho^2} \right) d\rho = \left[ \rho - \arctan \rho \right]_0^1 = 1 - \frac{\pi}{4}$$. - Multiply all: $$1 \times \frac{\pi}{2} \times \left(1 - \frac{\pi}{4} \right) = \frac{\pi}{2} \left(1 - \frac{\pi}{4} \right)$$. --- 4. **Part (b):** Find the volume bounded by $$x^2 + y^2 + z^2 = 4$$ (sphere radius 2), $$x^2 + y^2 + z^2 = 25$$ (sphere radius 5), and $$z = -\sqrt{\frac{1}{3} x^2 + \frac{1}{3} y^2}$$. - The volume is the region between two spheres and below the surface. - Rewrite the plane: $$z = -\sqrt{\frac{1}{3} (x^2 + y^2)} = -\frac{1}{\sqrt{3}} \sqrt{x^2 + y^2}$$. - In cylindrical coordinates: $$x = r \cos \theta$$, $$y = r \sin \theta$$, $$z = z$$. - The surface is: $$z = -\frac{r}{\sqrt{3}}$$. - The spheres are: $$r^2 + z^2 = 4$$ and $$r^2 + z^2 = 25$$. - The volume is between $$r^2 + z^2 = 4$$ and $$r^2 + z^2 = 25$$, and above $$z = -\frac{r}{\sqrt{3}}$$. - Use spherical coordinates: $$x = \rho \sin \phi \cos \theta$$, $$y = \rho \sin \phi \sin \theta$$, $$z = \rho \cos \phi$$. - The surface equation becomes: $$\rho \cos \phi = -\frac{\rho \sin \phi}{\sqrt{3}} \implies \cos \phi = -\frac{\sin \phi}{\sqrt{3}} \implies \tan \phi = -\sqrt{3}$$. - Since $$\phi \in [0, \pi]$$, $$\tan \phi = -\sqrt{3}$$ corresponds to $$\phi = \pi - \frac{\pi}{3} = \frac{2\pi}{3}$$. - So $$\phi$$ ranges from $$\frac{2\pi}{3}$$ to $$\pi$$ (below the cone). - The volume integral is: $$\int_0^{2\pi} \int_{2}^5 \int_{2\pi/3}^{\pi} \rho^2 \sin \phi \, d\phi \, d\rho \, d\theta$$. - Compute: $$\int_0^{2\pi} d\theta = 2\pi$$, $$\int_2^5 \rho^2 d\rho = \frac{5^3}{3} - \frac{2^3}{3} = \frac{125 - 8}{3} = \frac{117}{3} = 39$$, $$\int_{2\pi/3}^\pi \sin \phi \, d\phi = [-\cos \phi]_{2\pi/3}^\pi = (-\cos \pi) - (-\cos \frac{2\pi}{3}) = (1) - ( -(-\frac{1}{2})) = 1 - \frac{1}{2} = \frac{1}{2}$$. - Multiply all: $$2\pi \times 39 \times \frac{1}{2} = 39 \pi$$. - So the volume is $$39 \pi$$. --- 5. **Part (c):** Convert $$\int_{-1}^1 \int_0^{\sqrt{1-y^2}} \int_{x^2+y^2}^{\sqrt{x^2+y^2}} xyz \, dz \, dy \, dx$$ to cylindrical coordinates. - Cylindrical coordinates: $$x = r \cos \theta$$, $$y = r \sin \theta$$, $$z = z$$, with $$r \geq 0$$, $$\theta \in [0, 2\pi]$$. - The Jacobian is $$r$$. - The limits: $$x$$ from $$-1$$ to $$1$$, $$y$$ from $$0$$ to $$\sqrt{1 - x^2}$$ (upper half circle), $$z$$ from $$x^2 + y^2 = r^2$$ to $$\sqrt{x^2 + y^2} = r$$. - So $$z$$ from $$r^2$$ to $$r$$. - The region in $$xy$$-plane is the upper half unit disk. - In cylindrical coordinates: $$r \in [0,1]$$, $$\theta \in [0, \pi]$$ (upper half circle), $$z \in [r^2, r]$$. - The integrand: $$xyz = (r \cos \theta)(r \sin \theta) z = r^2 \cos \theta \sin \theta z$$. - The volume element: $$dz \, r dr \, d\theta$$. - The integral becomes: $$\int_0^{\pi} \int_0^1 \int_{r^2}^r r^2 \cos \theta \sin \theta z \, dz \, r dr \, d\theta = \int_0^{\pi} \int_0^1 \int_{r^2}^r r^3 \cos \theta \sin \theta z \, dz \, dr \, d\theta$$. - This is the converted integral in cylindrical coordinates. --- **Final answers:** (a) i. $$16 \int_0^2 e^{t^3} dt$$ (a) ii. $$\frac{\pi}{2} \left(1 - \frac{\pi}{4} \right)$$ (b) Volume = $$39 \pi$$ (c) Cylindrical integral: $$\int_0^{\pi} \int_0^1 \int_{r^2}^r r^3 \cos \theta \sin \theta z \, dz \, dr \, d\theta$$